Elements of
Fracture Mechanics
Elements of
Fracture Mechanics
Prashant Kumar
Former Professor
Department of Mechanical Engineering
IIT Kanpur
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To
Many excellent teachers who motivated me from my school
days to PhD program. Some of them are:
K. Kumar, Lucknow Montessori School, Lucknow (1956-57)
T. Joshi, Lucknow Christian College, Lucknow (1960-62)
N. C. Dahl, IIT Kanpur, Visiting from MIT, USA (1963-64)
A. J. Erickson, IIT Kanpur, Visiting from MIT, USA (1963-65)
P. W. Fay, IIT Kanpur, Visiting from Cal. Tech., USA (1965-66)
C. W. Radcliff, University of California, Berkeley, USA (1967-68)
Y. Takahashi, University of California, Berkeley, USA (1967-68)
R. J. Clifton, Brown University, USA (1970-77)
L. B. Freund, Brown University, USA (1971-74)
A.C. Pipkin, Brown University, USA (1972-73)
P.C. Paris, Brown University, Visiting from Lehigh University, USA (1973-74)
Barton Roessler, Brown University, USA (1973-74)
J.R. Rice, Brown University, USA (1972-73)
and
My uncle Dr. Ram Krishan and aunt Mrs. Sheela Krishan who raised me and played
the most pivotal role in shaping my thought process.
Preface
There are many components of machines, process plants and household goods that fail through
fatigue and fracture, which can be avoided by applying fracture mechanics. Although the modern
fracture mechanics was born in 1948–49, its acceptability was quite slow for several decades because
of its complex mathematics and involved concepts. Initially, the fracture mechanics was applied to
high-risk products like nuclear plants, airplanes, space vehicles, submarines, etc. Now this field is
becoming popular at the grassroot levels too. Therefore, it is important to explain this concept in
simple, well-disposed and easy-to-understand manner without compromising on the rigor—this
book is intended to do the same.
The book has gone through several phases of revisions to improve its readability further. Several
analogies and anecdotes placed appropriately in this book have also helped in accomplishing this
goal. Two most important mechanisms—fatigue failure and environment-assisted fracture that
result in the growth of subcritical cracks to catastrophic failure of structural components—have
been discussed in Chapter 9. Along with that, an additional chapter has talked about various nondestructive test methods for identifying cracks in the structural components. The detailed
illustrations developed with the help of latest graphic software have added more clarity in
understanding the concept.
This book has adopted the approach of different courses on fracture mechanics at the advanced
undergraduate and postgraduate levels. The readers novice in this field can explore this complex
but beautiful concept effortlessly through this book. The responses of students have directed me in
choosing the right pace and the appropriate levels while presenting the subject. Readers with prior
knowledge of fracture mechanics may find it rudimentary at few places. But with experience I have
learned to be on elementary side rather than leaving a reader confused.
Unlike in most books on fracture mechanics where basic concepts are discussed shortly usually
leaving the reader confused, this book gradually discusses the fundamental theory with the results
and their applications. With my industrial experience and interest in product design, I view the
theories of fracture mechanics as useful tools to be applied to practical problems. Once the
fundamentals of fracture mechanics are mastered properly, the reader can easily move to advanced
books or research journals. Therefore, special tricks involving highly complex mathematics to solve
problems of minor interest are not included here.
I hope this book will be appreciated by the readers for its salient features like lucid content,
articulation, anecdotes, and presentation of the subject in small and effective steps, and will prove
to be beneficial in introducing the subject in general.
PRASHANT KUMAR
Acknowledgements
• IIT Kanpur for encouragement and financial support to prepare the manuscript
• ARDB, Sena Bhavan, New Delhi for supporting various research grants in the field of fracture
mechanics
• Prof. N.N. Kishore (IIT Kanpur) for contributing a chapter in the book
• Prof. Raju Sethuraman (IIT Madras) for contributing a chapter in the book
• Prof. K. Ramesh (IIT Madras) for contributing a figure on photoelastic fringes
• Profs. Bishakh Bhattacharya, J. Ramkumar and K. K. Kar for constructive criticism of the
manuscript
• Raj Mulani, Rahul Ranjan Pandey, Ramesh Chandra and Yatendra for preparing the
illustrations using latest graphic softwares
• Vinay Pahlajani and Rajamanohar for their help in graphics through latest softwares
• Mr. Divakar to coordinate various activities in preparing the manuscript
• Mr. Anurag Goel to meet various needs of the manuscript preparation
• Mrs. Sandhya Agnihotri and S.L. Yadav for typing the manuscript
• Mr. Kalyan Kumar Singh for constructive suggestions and help in proof reading
• My son Saurabh Vishal on persistent encouragement to have the book published
• McGraw-Hill Education (India) Ltd. for publishing the book
Contents
Preface
Acknowledgements
vii
ix
1. Background
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1
Kinds of Failure 1
Historical Aspects 3
Brittle and Ductile Fracture 4
Modes of Fracture Failure 5
How Potent is a Crack? 6
Point of View 7
Damage Tolerance 7
References 7
2. Energy Release Rate
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
Introduction 9
Griffith’s Dilemma 9
Surface Energy 10
Griffith’s Realization 10
Griffith’s Analysis 11
Energy Release Rate 14
2.6.1 Definition 14
2.6.2 Mathematical Formulation 15
2.6.3 Change in Compliance Approach 16
2.6.4 Change in the Strain Energy Approach
Energy Release Rate of DCB Specimen 21
Anelastic Deformation at Crack-tip 24
Crack Resistance 25
Stable and Unstable Crack Growth 26
R-curve for Brittle Cracks 27
Thin Plate vs Thick Plate 28
Critical Energy Release Rate 29
Closure 31
Questions 31
Problems 32
References 34
9
20
xii
Contents
3. Stress Intensity Factor
3.1
3.2
3.3
3.4
3.5
3.6
Introduction 35
3.1.1 Why Should Investigations be Closer to the Crack Tip? 35
3.1.2 Linear Elastic Fracture Mechanics (LEFM) 35
Stress and Displacement Fields in Isotropic Elastic Materials 36
Stress Intensity Factor 38
Background for Mathematical Analysis 41
3.4.1 Field Equations 42
3.4.2 Elementary Properties of Complex Variables 45
Westergaard’s Approach 47
3.5.1 Mode I (Opening Mode) 47
3.5.2 Mode II (Sliding Mode) 56
3.5.3 Mode III (Tearing Mode) 58
Concluding Remarks 59
Questions 60
Problems 60
References 61
4. SIF of More Complex Cases
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
35
Other Applications of Westergaard Approach 62
4.1.1 Wedge Loads on Cracked Surfaces 62
4.1.2 Collinear Cracks in an Infinitely Long Strip 64
Application of the Principle of Superposition 66
4.2.1 Internal Pressure on Cracked Faces 67
4.2.2 Wedge Load at the Surface of a Crack Face 68
Crack in a Plate of Finite Dimensions 69
Edge Cracks 71
Embedded Cracks 72
4.5.1 Elliptical Crack 73
4.5.2 Semi-elliptical Cracks 74
4.5.3 Quarter or Corner Cracks 75
The Relation between GI and KI 75
Critical Stress Intensity Factor 78
Bending and Twisting of Cracked Plates 80
4.8.1 Terminology of the Plate Theory 80
4.8.2 Through-the-Thickness Crack in a Plate 81
4.8.3 Bending Moment on a Centre-Cracked Plate 82
Closure 84
APPENDIX 4A
General Approach to Determine Stress and
Displacement Fields 84
62
Contents
APPENDIX 4B
Questions 95
Problems 96
References 98
SIF of Some Important Cases
89
5. Anelastic Deformation at the Crack Tip
5.1
5.2
5.3
5.4
5.5
Further Investigation at the Crack Tip 99
Approximate Shape and Size of the Plastic Zone 100
5.2.1 Plastic Zone Shape for Plane Stress 101
5.2.2 Plastic Zone Shape for Plane Strain 103
Effective Crack Length 104
5.3.1 Approximate Approach 104
5.3.2 The Irwin Plastic Zone Correction 105
5.3.3 Plastic Zone Size through the Dugdale Approach
Effect of Plate Thickness 111
Closure 115
Questions 115
Problems 116
References 116
99
108
6. J-Integral
6.1
6.2
6.3
6.4
6.5
6.6
6.7
118
Relevance and Scope 118
Definition of the J-Integral 119
Path Independence 122
Stress-Strain Relation 125
Further Discussion on J-Integral 127
6.5.1 From a Designer’s Point of View 127
6.5.2 Experiments to Determine the Critical J-Integral 127
6.5.3 Comments on the Numerical Evaluation of J-Integral 128
6.5.4 Predicting Safety or Failure 128
6.5.5 Comments on the Experimental Determination of the Toughness of
Ductile Materials 129
Engineer Approach—A Short Cut 130
6.6.1 A Simplified Relation for the J-Integral 130
6.6.2 Applications to Engineering Problems 131
Closure 135
APPENDIX 6A Equivalence of G and J for Elastic Materials 136
APPENDIX 6B
Questions 145
Problems 145
References 147
xiii
The J-Integral of Some Common Cases through
Engineering Approach 138
xiv
Contents
7. Crack Tip Opening Displacement
7.1
7.2
7.3
7.4
149
Introduction 149
Relationship between CTOD, KI and GI for Small Scale Yielding 150
Equivalence between CTOD and J 151
Closure 152
Questions 153
Problems 153
References 153
8. Test Methods
8.1
8.2
8.3
8.4
8.5
8.6
154
Introduction 154
KIc-Test Technique 156
8.2.1 Various Test Specimens 156
8.2.2 Constraints on Specimen-Dimensions 158
8.2.3 A Dilemma 158
8.2.4 Fatigue Crack Growth to Sharpen the Tip 159
8.2.5 Clip Gauge 160
8.2.6 Load-Displacement Test 161
8.2.7 Measuring the Crack Length 162
8.2.8 Data Analysis 163
8.2.9 Comments on Plane Strain KIc–Test 164
Test Methods to Determine JIc 164
8.3.1 Graphical Interpretation 165
8.3.2 Historical Development 166
8.3.3 Formulation 166
8.3.4 Details of JIc Test Method 168
8.3.5 Comments on JIc–Test 171
Test Methods to Determine GIc and GIIc 172
8.4.1 Determination of Interlaminar GIc 172
8.4.2 Determination of Interlaminar GIIc 177
Determination of Critical CTOD 179
Closure 183
APPENDIX 8A
Questions 183
Problems 184
References 187
Regression Analysis to Fit a Line of Known Slope
9. Fatigue Failure and Environment-assisted Fracture
9.1
9.2
Introduction 188
Fatigue Failure 188
183
188
Contents
9.3
9.4
9.5
9.2.1 Terminology 189
9.2.2 S-N Curve 191
9.2.3 Crack Initiation 192
9.2.4 Crack Propagation 195
9.2.5 Effect of an Overload 198
9.2.6 Crack Closure 199
9.2.7 Variable Amplitude Fatigue Load 201
Environment-assisted Fracture 202
9.3.1 Introduction 202
9.3.2 Micromechanisms 203
9.3.3 Test Methods 205
9.3.4 Major Factors Influencing Environment-assisted Fracture
9.3.5 Liquid Metal Embrittlement 210
9.3.6 Design Considerations 210
Environment-assisted Fatigue Failure 211
Closure 213
Questions 214
Problems 214
References 216
10. Finite Element Analysis of Cracks in Solids
xv
209
218
10.1 Finite Element Method 218
10.2 Direct Methods to Determine Fracture Parameters 223
10.3 Indirect Methods to Determine Fracture Parameters 226
10.3.1 J-Integral Method 226
10.3.2 Energy Release Rate Method 227
10.3.3 Stiffness Derivative Method 228
10.3.4 Singular Element Method 228
10.3.5 Barsoum Element 230
References 232
11. Mixed Mode Crack Initiation and Growth
11.1 Introduction 233
11.2 Fracture Surface 233
11.3 Mixed Mode Crack Propagation Criteria 234
11.3.1 Modified Griffith Criterion 235
11.3.2 Maximum Tangential Stress Criterion 236
11.3.3 Strain Energy Density Criterion 240
11.4 An Example of Mixed Mode 244
11.5 Crack Growth 247
References 247
233
xvi
Contents
12. Crack Detection through Non-Destructive Testing
248
12.1 Introduction 248
12.2 Examination through Human Senses 249
12.2.1 Visual Inspection 249
12.2.2 Investigation through Hearing 250
12.2.3 Detection through Smell 250
12.2.4 Other Simple Methods 250
12.3 Liquid Penetration Inspection 251
12.3.1 Principle 251
12.3.2 Procedure 252
12.3.3 Crack Observation 252
12.4 Ultrasonic Testing 253
12.4.1 Principle 253
12.4.2 Equipment 255
12.4.3 Immersion Inspection 258
12.5 Radiographic Imaging 259
12.5.1 Contrast through Absorption Rate 259
12.5.2 Imaging through X-rays 259
12.5.3 Imaging through Gamma Rays 261
12.5.4 Strong Points of Radiographic Imaging 262
12.5.5 Limitations of Radiographic Imaging 262
12.6 Magnetic Particle Inspection 263
12.6.1 Principle 263
12.6.2 Sensitivity 263
12.6.3 Hardware 264
12.6.4 Flaw Orientation 264
12.6.5 Magnetic Ink Powder 265
12.6.6 Voltage Source 265
12.6.7 Demagnetization 265
12.6.8 Strength and Limitations 266
12.7 Concluding Remarks 266
References 267
Index
269
Chapter
1
Background
I learned there that innovation is a very difficult thing in the real world.
R.P. Feynman
1.1 KINDS OF FAILURE
Some people suffer from hypertension and are susceptible to heart attacks. Many, unfortunately,
suffer from diabetes and take special precautions to avoid or delay its destructive effects. On an
average, a person in our society is vulnerable to one or two of such diseases as cancer, arthritis,
asthma, hepatitis, gastritis, tuberculosis, etc. and is conscious about it. Similarly, a component in
a structure may be susceptible to one, two or more kinds of failure. For example, under given
load conditions a roller bearing is most likely to fail through fatigue of its rollers after a certain
number of rotations.
We should thus know the different conditions that can cause the failure of a structural
component. Some of the common causes of failure are:
∑ Yielding
∑ Deflection beyond a certain stage
∑ Buckling
∑ Fatigue
∑ Fracture
∑ Creep
∑ Environmental degradation
∑ Resonance
∑ Impact
∑ Wear
A component is designed so as to avoid yielding of the worst loaded point (critical point).
Safety against yield failure is considered to be the basic requirement of any design and is taught
in all courses on ‘strength of materials’ at undergraduate level. Although stress tensor is quite
complex with six independent components, criteria like Mises or Tresca are adopted to obtain a
2
Elements of Fracture Mechanics
scalar number (e.g., maximum shear stress in Tresca criterion). The scalar number is then compared
with a limiting value which is determined through experiments. In simple cases, the worst loaded
point may easily be identified, whereas it may be difficult to find where exactly the worst loaded
point is in a general case. These days excellent computer packages based on finite element analysis
are available. In a finite element simulation, the body of the component can be divided into
different colors; for example, red color shows the portion where the stresses are maximum, green
color shows the lowest stresses and other colors for intermediate stresses. However, it has been
found that often a structural component fails even when the worst loaded point is well within the
yield stress. Thus we conclude that a design of a work-component based entirely on avoiding
yielding is not adequate in certain cases. The component may be susceptible to crack growth!
All engineering materials deform on loading. A structural component may be deformed to
such an extent that its performance is affected. Certain plastics have been developed to possess
high strength, but they may not be suitable for many structural components because of their low
stiffness, i.e., about 1% of steel’s stiffness. If the wings of an airplane are made of a polymer, they
may droop down so much that the tip of the wings would touch the ground. In fact, many
components of conventional engineering materials (steel, aluminum, and the like) are designed
to meet constraints of deflection, although the stress of the worst loaded point is considerably
lower than yield stress. These kinds of components are thus susceptible to failure through
deflection beyond a certain value. If a crack grows in a component, its stiffness decreases and the
deformation may exceed the allowable limit.
A thin member under a compressive load or a thin tube under torsion or lateral compression
may be susceptible to buckling, and the design should be checked against likely failure through
buckling.
Many components of the modern industrial world are subjected to fluctuating loads and
consequently may fail through fatigue. In fact, failure through fatigue is so common that more than
80% failures are caused by fluctuating loads. Many investigators are currently working for the
development of this field. However, convenient and effective methods to control fatigue-failures are
still not adequately developed. A critical structural component should be regularly checked to detect
fatigue cracks through non-destructive tests. This has led to the development of excellent methods of
crack identification, such as ultrasonic crack detection, X-ray or radiation filming, detecting through
monitoring acoustic emission, magnetic flux method, decoration of surface cracks through dyepenetration, etc.
Fracture mechanics is based on the implicit assumption that there exists a crack in a workcomponent. The crack may be man-made such as a hole, a notch, a slot, a re-entrant corner, etc.
The crack may exist within a component due to manufacturing defects like slag inclusion, cracks
in a weldment or heat affected zones due to uneven cooling and presence of foreign particles. A
dangerous crack may be nucleated and grown during the service of the component (fatigue
generated cracks, nucleation of notches due to environmental dissolution). Fracture mechanics
deals with the question—is a known crack likely to grow under a certain given loading condition?
Fracture mechanics is also applied to crack growth under fatigue loading. Initially, the fluctuating
load nucleates a crack, which then grows slowly and finally the crack growth rate per cycle picks
up speed. Thereafter comes the stage when the crack-length is long enough to be considered
critical for a catastrophic fracture failure.
About 50–60 years ago, when accurate analysis for predicting the growth of a crack was not
available, a reasonably high factor of safety was chosen to account for unforeseen factors. A large
Background
3
part of this ambiguity has been cleared with the development of fracture mechanics and
understanding the causes and effects of fatigue failure. This now enables a designer to use a
much lower factor of safety, thus reducing cost of such structural components. Simultaneously,
the weight of these components is reduced and their reliability is enhanced.
Other ways of failure (creep, environmental degradation, wear, etc.) are also as important and
must be looked into and analyzed, so that the component may not develop snags on their account.
However, one thing should also be borne in mind, that a component is usually not likely to fail
through more than two or three ways. Therefore, susceptibility should be kept in mind at the time
of designing a component.
1.2 HISTORICAL ASPECTS
For a long time, we always had some idea about the role of a crack or a notch. While cutting a tree,
we would make a notch with an axe at its trunk and then pull it down with a rope. While breaking
a stick we would make a small notch with a knife before bending it. Swords played an important
role in the pre-industrial society. Good swords were made by folding a thin metal sheet at the
centre line and then hammering it to make it thin again so that it could be further folded. Thus, a
sword would have many layers. If a crack develops in one of the layers, it is not likely to move to
another, thus making the sword very tough.
Leonardo da Vinci (1452–1519) was the first person to make a setup to measure the strength of a
wire. He found out that the strength of a wire depended on its length. A wire in his time were not of
high quality and a long wire was likely to posses many cracks. However, fracture mechanics was
not studied as a separate discipline for a long time.
The Industrial Revolution opened a new vista for us and many different kinds of machines
and structures were designed and built. If we look back and study, we would find that many
bridges, boilers, buildings, ships, were failed due to fracture in nineteenth century [1.1].
Locomotives, a very important industry in those days, used to face innumerable accidents due to
failure of wheels, axles of wheels, and rails. Wohler [1.2] was one of the earliest investigators
(1860), who investigated fatigue of locomotive axles by applying controlled cyclic loads. This led
to development of S-N Diagram and finding endurance limit of steel [1.3].
In nineteenth century and early part of twentieth century, the entire industry was obsessed
with production. Even the failure theories were developed quite late: Tresca in (1864) and Mises
(1913). However, World War II accelerated the industrial production at a very rapid rate, due to
unusually high demands of war. Within six years of the war, the know-how of aircraft making
improved dramatically. Also, the ships, which were earlier made by joining plates together through
the process of riveting, were changed to welded frames. Many cargo ships, known as liberty
ships, were rolled out from American docks within a short span. However, soon there were
complications regarding welded structures. Many of these failed in the cold temperatures of the
North Atlantic Ocean [1.1]. As a matter of fact some of these broke up into two parts. However,
ships made by riveting plates did not display such failures. If a crack nucleated and grew in a
plate, it would only split that plate into two parts; the crack would not grow into another plate. A
welded structure is a large single continuous part and therefore, if the crack becomes critical, it
will run through the entire hull of the ship.
4
Elements of Fracture Mechanics
Jet planes were developed initially as small fighter planes, which would fly at high altitudes.
When this technology was used for the construction of large body passenger planes, it was found
that the planes were exploded in the air. An investigating team was appointed to look into the
failures of the Comet Jetliner right in the early fifties. It was found that a fatigue crack, initiated
near an opening in the fuselage [1.4], ran through its entire body especially at high altitudes,
where the outside atmospheric pressure was low and the interior of the airplane was pressurized
for the comfort of passengers. In fact, a jetliner flying over a height of 10,000 m (32,800 ft) works
like a pressurized balloon with the wall of its fuselage under high tensile stresses.
With the development of large ships made of welding plates and high capacity jet airplanes,
new problems arose; the predominant question asked was—what causes the failure? Can we
contain the failure? And then a new discipline of engineering, called ‘fracture mechanics’ was
developed. In fact, Griffith [1.5, 1.6] developed the right ideas for the growth of a crack in the
1920s. Using atomistic models, he estimated the strength of a structural material and found that
it should be of the order of its modulus. However, the strength of engineering materials was
experimentally found to be of two to three lower orders. He went ahead and developed the ideas
of energy requirements in growing a crack. However, his work was not taken seriously at that
time because engineers were busy in increasing production and making money. This period was
followed by the Great Depression which had many problems of its own. Further, Griffith was not
able to invent a convenient parameter to predict the load on a component that would cause the
growth of a crack.
For all practical purposes, modern fracture mechanics was born in 1948 when George Irwin [1.7]
formulated fracture mechanics by devising workable parameters like stress intensity factor and
energy release rate. Once the breakthrough took place, many investigators started taking interest in
it and fracture mechanics became a separate and important discipline with several reputed journals
and text books. Irwin’s development was mainly for brittle or less ductile materials. The analysis
was conservative for most engineering materials which are generally ductile. Other parameters,
like Crack Tip Opening Displacement by Wells [1.8] in 1961 and J-Integral by Rice [1.9] in 1968,
were developed to account for the large plastic zone at the crack-tip in ductile materials.
Fracture mechanics is now applied extensively to important fields like nuclear engineering,
piping, space ships, rockets, offshore structures, etc. Critical components in nuclear power plants
are made from very tough materials; but they too have failed catastrophically once in a while.
1.3 BRITTLE
AND
D UCTILE FRACTURE
Some materials are known as brittle because a crack moves easily through components made of
such materials. If we investigate a fractured surface of a brittle failure to determine the depth up
to which the material is affected by the crack growth, we find that material was influenced to a
very shallow depth. Rest of the material remains unaffected. On the contrary, a ductile fracture
causes a large amount of plastic deformation to a significant depth.
Brittle fracture in crystalline metals can be classified into two broad groups—intergranular
and transgranular. A crack tip of intergranular failure grows along the grain boundaries as shown
in [Fig. 1.1 (a)]. Transgranular fracture, on the other hand, occurs through the crack tip propagating
within grains [Fig. 1.1 (b)]. However, cleavage failure within a grain occurs along a weak
crystallographic plane. In fact, cleavage fracture is the most brittle form of a fracture and it hardly
Background
5
damages the fractured surfaces. Once the cleavage crack reaches the grain boundary, it finds
another favorable orientation in the next grain.
(a) Intergranular
(b) Transgranular
Fig. 1.1 Brittle fracture
Ductile fracture growth occurs due to substantial plastic deformation and creation of microvoids
in the vicinity of the crack tip. The material deforms plastically due to micromechanisms, such as
nucleation and motion of dislocations, formation of twins, etc. Engineering materials generally
contain second phase particles. Tiny voids are formed at the sides of these particles under the
influence of the tensile field of the crack tip. Dislocation motion helps in the formation of these
voids. The ductile crack growth occurs by the coalescence of these voids. Fractured surface of a
ductile failure shows tiny dimples and gives the surface a rather rough look. In fact, around one
such dimple, a second phase particle can be identified. The plastic deformation and coalescence
of voids absorb a large amount of energy and, therefore, a crack does not grow easily in ductile
materials.
Often it has been found that materials normally ductile at room temperature in ordinary
conditions behave as brittle materials under certain special conditions. Steel, which is quite ductile
at room temperature, becomes brittle at low temperatures. This explains why welded structures
of Liberty ships in World War II failed in the cold waters of the North Atlantic Ocean. Also, the
toughness of certain materials is affected considerably by the rate of loading (strain rate).
A thick plate of a regular ductile material may also allow the growth of a crack in a brittle
manner. The portion that is deep inside the thick plate (away from free surfaces) is constrained
from all sides and large plastic deformations are not possible in the vicinity of the crack-tip. In
comparison to thick plates, thin plates are more resistant to crack growth. These aspects will be
discussed in detail in subsequent chapters.
1.4 MODES OF F RACTURE FAILURE
A crack front in a structural component is a line usually of varying curvature. Thus, the state of
stress in the vicinity of the crack front varies from one point to another. A segment of the crack front
can be divided into three basic modes as shown in Fig. 1.2. Mode I is the opening mode and the
displacement is normal to the crack surface. Mode II is a sliding mode and the displacement is in
the plane of the plate—the separation is antisymmetric and the relative displacement is normal to
the crack front. Mode III also causes sliding motion but the displacement is parallel to the crack
front, thereby causing tearing.
6
Elements of Fracture Mechanics
P
Crack front
Mode I
(opening mode)
P
Crack front
Crack front
P
P
P
Mode II
(sliding mode)
P
Mode III
(tearing mode)
Fig. 1.2 The three modes of fracture
An inclined crack front in a component can be modeled as a superposition of the three basic
modes and then, the effect of loading by each mode can be analyzed separately. Mode I usually
plays a dominant role in many engineering applications and is considered to be the most
dangerous. However, in certain applications, components fail through the dominant roles played
by Mode II or Mode III. Mode I has been analyzed most so far. Also, elaborate experimental
methods have been developed to determine toughness in Mode I; in fact, detailed codes have
been prepared for these experimental methods and they are internationally accepted.
1.5 HOW POTENT IS A CRACK?
Designers are always interested to know whether a crack is likely to grow if the geometry of a
crack in a structural component, loads and other boundary conditions are known. We should,
therefore, have a parameter to measure crack potency or crack extension force.
The analysis of fracture mechanic problem is done through different approaches, each having its
own parameter [1.10–1.12]. All of them are well accepted to measure the potency of a crack; only
one is needed to solve a problem. The parameter Energy Release Rate (G) is energy based and is
applied to brittle or less ductile materials. Stress Intensity Factor (K) is stress based, also developed
for brittle or less ductile materials. J-Integral ( J) has been developed to deal with ductile materials.
Its formulation is quite general and can be applied to brittle materials also. Crack Tip Opening
Displacement (CTOD) parameter has been also developed for ductile materials and, as the name
suggests, it is displacement based.
Background
7
Depending upon the application, an appropriate parameter is chosen to analyze the given
problem. For critical components of nuclear reactors, airplanes, submarines, locomotives, turbine
blades, etc., a sophisticated fracture mechanics analysis is recommended and one may gain by
investing more money for accurate analysis and experimental tests. In this book, all four
parameters are developed and discussed.
1.6 POINT OF VIEW
Fracture mechanics problems are studied through two different points of views: (a) material
science and (b) applied mechanics. Material scientists like to study microscopic mechanisms near
the crack front, such as dislocation generation and motion, role of grain boundaries, formulation
of twins, role of second phase particles, nucleation and growth of voids and their coalescence,
etc. They also study texture of the fractured surfaces. In this short book, we have taken the applied
mechanics approach and the material is assumed to be continuous, ahead of the crack-tip.
However, some discussion on micromechanisms has been included at a few places.
1.7 DAMAGE TOLERANCE
Problems of fracture mechanics are solved using two different approaches. In the first approach,
component geometry which includes the length, location and orientation of the crack is given
along with boundary conditions. The objective is to find the upper limit of the applied load that
would not cause catastrophic failure of the component.
In the second approach, known as damage tolerance, the maximum load on a component is
known; the objective is to find the longest length of a crack that remains dormant. Once we know
the length, the structural component can be thoroughly checked with an appropriate nondestructive test. In the case of fluctuating loads applied on the component, a fatigue crack may be
nucleated even at a surface which was previously crack free. This crack may grow with fluctuating
loads. In such situations, critical components are checked regularly. If a crack that is likely to
grow and become critical is detected, then the component is repaired or replaced. On the other
hand, detection of a small crack should not cause panic because its length may be much smaller
than the maximum length of crack allowed in damage tolerance analysis.
These days many companies have started believing in avoiding a likely catastrophic failure by
regular non-destructive tests of critical components. For example, a chemical company making
urea in Kanpur city maintains an excellent nondestructive test department. Its engineers mostly
face problems at the pipe-joints and thus they regularly check the joints, identify cracks and take
necessary actions. In the long run, it saves considerable expenses because a catastrophic failure
through the growth of a crack may cause extensive damage to other parts, besides causing shutdown of the plant and loss of human lives in some cases.
REFERENCES
1.1 Broek, D. (1982). Elementary Engineering Fracture Mechanics, Martinus Nijhoff Publishers,
The Hague.
8
Elements of Fracture Mechanics
1.2 Wöhler, A. (1860). Versuche über die Festigkeit der Eisenbahnwagenachsen Zeitschrift für
Bauwesen. 10; English summary (1967), Engineering, 4, pp. 160–161.
1.3 Goodman, J. (1899). Mechanics Applied to Engineering, London: Longmans Green.
1.4 Barsom, J.M. and Rolfe, S.T. (1987). Fracture and Fatigue Control in Structures, Prentice Hall,
Inc., Englewood Cliffs, New Jersey.
1.5 Griffith, A.A. (1921). The Phenomena of Rupture and Flow in Solids, Philosophical
Transactions of the Royal Society of London, A 221, pp. 163–197.
1.6 Griffith, A.A. (1924). The Theory of Rupture, Proceedings of the First International Conference
of Applied Mechanics, Delft.
1.7 Irwin, G.R. (1948). Fracture Dynamics, Fracturing of Metals, American Society for Metals,
Cleveland, pp. 147–166.
1.8 Wells, A.A. (1961). Unstable Crack Propagation in Metals: Cleavage and Fracture,
Proceedings of the Crack Propagation Symposium, College of Aeronautics, Cranfield, 1.
pp. 210–230.
1.9 Rice, J.R. (1968). A Path Independent Integral and the Approximate Analysis of Strain
Concentration by Notches and Cracks, Journal of Applied Mechanics, Transactions of ASME,
35, pp. 379–386.
1.10 Ramesh, K. (2007). E-Book on Engineering Fracture Mechanics, IIT Madras,
URL: http//apm.iitm.ac.in/smlab/kramesh/book_4.htm.
1.11 Liu, A. (2005). Mechanics and Mechanisms of Fracture—An Introduction, ASM International,
Ohio, USA.
1.12 Anderson, T.L. (2004). Fracture Mechanics: Fundamentals and Applications, CRC, Press-Book.
Chapter
2
Energy Release Rate
A man in passion rides a horse that runs away with him.
C.H. Spurgeon
2.1 INTRODUCTION
Whether a crack in a component is likely to grow under given load conditions is of vital importance
to fracture mechanics.
The problem has been analyzed through several approaches—stress, displacement or energy
methods [2.1–2.7]. Each approach defines a suitable parameter. A limit on the parameter defines
the toughness of the material. For a prescribed load condition, if the value of the parameter exceeds
the limit, the crack may grow.
This chapter deals with the energy method. The advantage of this energy approach is that
there is no need to account for the large stresses that are developed in the vicinity of a crack-tip.
The energy method conveniently avoids any analysis close to the crack-tip. The approach is similar
to solving the problem of a body sliding down on a frictionless slope in which one is interested to
know only the velocity of the body. It is determined by invoking the conservation of kinetic and
potential energies, irrespective of the slope of the slide, which may vary from one point to another
along the path.
2.2 GRIFFITH’S DILEMMA
Griffith, in the early 1920s [2.8, 2.9], developed some basic concepts. He was aware of the analytical
solution, developed by Inglis [2.10], which determines stress field around an elliptical hole in a
large plate, loaded under the tensile stress s 0 as shown in Fig. 2.1. He noted that the maximum
stress develops at point A of the ellipse and is given by
2a ˆ
Ê
max
s 22
= s 0 Á1 + ˜
Ë
b¯
(2.1)
10
Elements of Fracture Mechanics
Fig. 2.1 An elliptical hole in a stretched plate
max
For a circular hole, s 22
is three times greater than s 0. But for an elliptical hole where the
max
major axis a is much longer than the minor axis b, s 22
becomes quite large; so much so that for
a sharp crack with minor axis tending to be very small (of the order of interatomic distances), no
real material can sustain the stress. Thus, even for a small applied stress s 0, Eq. (2.1) suggests that
max
s 22
would be very large and would exceed the ultimate strength of the material. Equation (2.1)
further suggests that even a sharp crack of small length may grow and break the component into
two pieces. However, this is contrary to our observations. Griffith thus concluded that some other
mechanisms must be existing which helped materials to sustain solid forms.
2.3 SURFACE ENERGY
Similar to the surface tension of a liquid, all solid surfaces are associated with surface energies or
free energies. These energies are developed because atoms close to a surface behave differently
from atoms at the interior of the solid. The interior atoms are attracted or repulsed by the
neighboring atoms more or less uniformly from all directions. On the contrary, an atom on the
free surface has no neighboring atoms towards the exterior side of the surface, thus resulting in a
different equilibrium. In fact, atoms at the surface, as well as atoms just under them, have to
re-adjust to form the equilibrium and this develops a strain in the material close to the free surface.
Such surface deformation requires energy and is known as surface energy.
2.4 GRIFFITH’S R EALIZATION
Griffith realized that a crack in a body would not grow unless energy was released to overcome
the energy needs of forming two new surfaces, one below and one above the crack plane. The
surface energy of a material depends on the material properties. However, its magnitude is rather
small, of the order of 1 J/m2. Table 2.1 lists the surface energy of some of commonly encountered
solids.
Surface energy of the order of 1 J/m2 is considered insignificant (1 J energy will raise the
temperature of one teaspoon of water by 0.05o C only). In brittle materials such as silica, glass and
diamond, advancing cracks require small energies of the order of surface energies, and, therefore,
Energy Release Rate
11
once a crack starts advancing, it runs through the body easily, causing catastrophic failure. But
some additional mechanisms operate on most materials which do not allow cracks to grow at low
energies. These mechanisms will be discussed subsequently in this chapter.
TABLE 2.1 Surface energy of some commonly used materials
Surface Energy (J/m2)
Material
Copper
Mild Steel
Aluminum
NaCl
MgO
Glass pane
Ice
Diamond
0.98
1.20
0.60
1.35
3.30
2.30
0.07
5.50
2.5 GRIFFITH’S A NALYSIS
Let us consider a plate with no prior crack [Fig. 2.2(a)]. It is pulled and then maintained in tension
between two rigid supports [Fig. 2.2(b)]. Now, with a knife, a crack is cut at the centre of the plate
with the crack plane normal to the tensile stress. The crack length is increased gradually with the
help of the knife. A critical stage reaches when the crack starts growing on its own; i.e., without
any further need of the knife. How long is this critical length? How to predict it?
l2a
2a
(a)
(b)
(c)
Fig. 2.2 (a) A unstretched plate, (b) the stretched plate, and (c) introducing
a crack at the center
Before we answer these questions, we would like to point out that the plate becomes less stiff
as the crack advances. Consequently, for this case of the plate with ends held rigidly, the stress
within the plate decreases and the strain energy stored in the plate is reduced. The energy thus
released is available for the crack to grow.
12
Elements of Fracture Mechanics
To convince ourselves, a simple experiment may be conducted by taking a discarded tube of a
bicycle wheel and cutting it along the length to obtain a plate sheet. The sheet is stretched and
mounted on a frame. We can even use the arm-rests of a chair for the purpose. A small crack is
introduced with a blade or knife normal to the stretching direction and the crack is enlarged
gradually. At the critical length, the crack starts advancing slowly on its own and the knife is
withdrawn. The crack slowly picks up the speed and runs all the way to the side edges, thus,
snapping the rubber sheet into two pieces.
Finding a rigorous solution to the problem shown in Fig. 2.2(c) is a difficult task at this stage.
To understand Giriffith’s analysis, we will carry an approximate analysis in this section. The
plate is chosen to have its dimensions much larger than the longest crack to be considered. Then,
the stress at points far away from the crack is assumed to remain constant.
Most of the energy release, as the crack advances, comes from those parts of the plate which
are adjacent to the cracked surfaces, because they are traction free. For the sake of convenience,
the major area of the plate where its strain energy is released may be taken as a triangle on each
side of the crack plane as shown in Fig. 2.2 (c). In fact, other shapes such as a parabola will serve
the purpose; we have chosen it to be triangular to keep the algebra simple. With the increase in
crack length the base and the height of both triangles increase and, therefore, the area from which
the strain energy is released is proportional to the square of the crack length.
The height of a triangle is l(2a) where l is a constant. Then, the total release of energy ER is
determined by multiplying the area of both triangles with plate thickness B and the strain energy
density s 2/2E where s is the tensile stress and E is the Young’s modulus. Thus, released energy
is given by
Ês 2 ˆ
ER = (Volume of triangles) ¥ Á ˜
Ë 2E ¯
Ês 2 ˆ
2l a2 Bs 2
Ê1
ˆ
= 2 Á (2 a)(2l a)B˜ ¥ Á ˜ =
Ë2
¯
E
Ë 2E ¯
Rigorous analysis (Chapter 4, Problem 8) shows that l = p/2 for thin plates (plane stress) giving
p a2 Bs 2
(2.2)
E
Energy is required to create the two new surfaces. If g is the surface energy per unit area of one
surface, the surface energy required E S is
ER =
ES = 2(2a)Bg = 4aBg
(2.3)
The relations of Equations (2.2) and (2.3) are shown graphically in Figure 2.3. ER increases
parabolically whereas ES increases linearly with increasing crack length a. Consider a small crack
length, 2a0 whose length is incremented by Da. The slope of ER is smaller than the corresponding
slope ES and, therefore, the energy release DER in advancing the crack-tip by distance Da is not
sufficient to meet the energy needs of new surfaces (DES). The crack would not grow and would
remain subcritical. In fact, the crack would be dormant unless energy is supplied by an external
agency, the operator of the knife in this case.
As the operator gradually continues cutting the crack further, the slope of ER increases, while
the slope of ES remains the same. A stage is reached when the slope of ER becomes equal to the
Energy Release Rate
13
slope of ES. The crack now becomes critical, because for an incremental advance of crack length,
energy release equals the energy required. Therefore, for the crack to become critical
dER
dES
≥
(2.4)
da
da
A question may be raised here—why does the crack grow in this experiment even when ER < ES?
Note that the difference in these two energies has already been supplied by the operator of the
knife. In fact, this difference is an extremely useful quantity because of which many small cracks
in a body do not grow and they remain dormant.
If the plate had been pulled to a higher tension prior to the introduction of the crack, the ER
would increase with the crack length at a faster rate (Fig. 2.3) and then even the cracks of smaller
lengths would become critical. In order to determine the critical crack length ac we substitute ER
[Eq. (2.2)] and ES [Eq. (2.3)] in Inequality 2.4 to obtain
E R , ES
2p ac Bs 2
≥ 4Bg
E
ES
ER
DES
DER
0
Da
0
a0
a
ac
Fig. 2.3 Variation of energy release ER and required surface
energy ES with crack length
For a safe crack,
ac £
2Eg
ps 2
(2.5)
14
Elements of Fracture Mechanics
If we want to know how much stress is required to advance a given crack for plane stress cases,
the inequality is rearranged to
1/2
È 2Eg ˘
sc ≥ Í
Î p a ˙˚
For plane strain (thick plate), E is replaced by E/(1–v2) and the relation becomes
È 2Eg
˘
sc ≥ Í
2 ˙
ÍÎ p (1 – v )a ˙˚
(2.6)
1/2
(2.7)
where v is the Poisson’s Ratio.
We deduce, from Inequalities 2.6 and 2.7, that the critical stress depends on modulus E, surface
energy g and crack length a. As expected, the higher value of the surface energy of a material
increases the critical stress whereas a longer crack reduces it. Larger modulus means that the plate
is capable of storing less energy, thereby resulting into smaller energy release which, in turn requires
higher stress for making the crack critical. One may also note from Inequality 2.6 that the product
sc a depends only on material properties (elastic constants E and v and surface energy g ). Therefore,
s a may be treated as a new variable in fracture mechanics (stress intensity factor) to be
developed in Chapter 3.
Griffith acknowledged with developing the correct theses in the 1920s but his analysis was not
developed to the extent that a designer can employ it to solve practical problems. For example, a
reader interested in the history of science knows that Kepler had almost developed everything
about the laws of motion, but they were not simple enough for a designer to make a machine.
The laws of motion were exploited only after Newton put them into a simple form with variables
clearly defined. In case of fracture mechanics also, it took another two and a half decades for
Irwin [2.11] and Orowan [2.12] to polish Griffith’s concepts and to define a variable, energy release
rate, which could be understood more easily and measured experimentally.
2.6 ENERGY R ELEASE R ATE
2.6.1 Definition
As discussed in the previous section, two important quantities are invoked— (i) how much energy
is released when a crack advances and (ii) minimum energy required for the crack to advance in
forming two new surfaces. The first quantity is measured with a parameter, energy release rate,
denoted by the symbol G after Griffith. The energy release rate is defined as energy release per
unit increase in area during crack growth. The word “rate” is sometimes confusing to beginners
because in most engineering applications, rate signifies differentiation with respect to time. In
the definition of G the rate is defined with respect to change in crack area. Another aspect of the
definition is that the energy release rate can be calculated even for the cracks which cannot grow
under a given load condition. That is, if there is a virtual growth of the crack, energy equal to G
would be released from the system per unit extension of area.
The energy requirement for a crack to grow per unit area extension is called crack resistance
and is usually denoted by the symbol, R. Note that symbol R is used in place of surface energy
g discussed in the previous section because during crack growth an anelastic deformation
Energy Release Rate
15
(e.g., plastic deformation in metals) also occurs up to a certain depth of the cracked surfaces. R, in
fact, is the sum of the energies required, (i) to form two new surfaces and (ii) to cause anelastic
deformation. Like energy release rate, the crack resistance is also a rate but it is rather unfortunate
that the word ‘rate’ has not been included in its nomenclature.
Both parameters, energy release rate as well as crack resistance, are important to study the
possibility of a crack becoming critical. Obviously, the energy release rate of a crack must be
greater than the crack resistance for the crack to have a chance to grow. In fact, this is not the only
condition; other conditions will be discussed subsequently. We would like to compare the growth
of crack with a young man trying to purchase a car. If he does not have enough money, he cannot
own the car. If he has just enough money to buy the car, he will make the purchase and bring the
car home. In case, he has money in excess, not only he will purchase the car, but he can drive
around at fast speeds to far away places and may even break road regulations and be dangerous
to other vehicles. Similarly, if the energy release rate exceeds the crack resistance, the crack acquires
kinetic energy and may grow at a speed faster than the speed of a supersonic airplane.
2.6.2 Mathematical Formulation
With an advancing crack, the following happens in a general case:
1. Strain energy in the component decreases or increases.
2. Stiffness of the component decreases.
3. The points of the component, at which external loads are applied, may or may not move.
Work is being done on the component by these forces if the points move.
4. Energy is being consumed to create two new surfaces.
Formulation for energy release rate is carried out by invoking the conservation of energy. Consider
the case of an incremental increase in the crack area DA. To cause this crack growth, an incremental
external work DWext is done by the external forces and the strain energy within the body of the
component increases by DU. Then, the available energy, GDA, provides the energy balance as
follows:
GDA = GDWext – DU
(2.8a)
Dividing the equation by DA and taking the limit DA Æ 0 , we obtain
d
(U – Wext)
(2.8b)
dA
The negative sign has been deliberately taken out of the differential term because (U – Wext) is
commonly known as potential energy P. Therefore, the equation is written as
G=–
dP
(2.9)
dA
The above equation is quite powerful to evaluate the energy release rate of a system. The equation
may be viewed from a different angle; that is, energy is available from the system if the potential
energy decreases. Note that G is always positive for a crack studied for its probable growth.
In many engineering applications, fracture mechanics is applied to plates of uniform thickness
and then DA can be expressed as BDa, where B is the thickness and Da is the increment in crack
length. Equation (2.9) is then modified to
G=–
16
Elements of Fracture Mechanics
1 dP
(2.10)
B da
Before we show an application of the equation, we would like to modify it for dynamic crack
propagation. As a crack moves rapidly, some energy is being consumed to the impart kinetic
energy to cracked portions of the body and to generate stress waves. Therefore, Eq. (2.8a) is
modified to
G=–
GDA = DWext – DU – DT
where DT is the incremental increase in kinetic energy in the body. On taking the limit, the equation
becomes
d
dT
(U – DWext) –
(2.11)
dA
dA
Equation (2.9) looks simple but one still has to worry for external forces and change in internal
energy. Can we simplify the equation further? For some cases we can. Two approaches are
developed: (i) change in compliance, (ii) change in internal energy.
G=–
2.6.3 Change in Compliance Approach
In a component, decrease in stiffness with an increasing crack length may be simple to visualize,
but in fracture mechanics, it is easier to deal with compliance which is the inverse of stiffness.
Thus, compliance of a body increases with the increase in the crack length.
Consider a general case of a body with a crack and load P as shown in Fig. 2.4. The displacement
u of the point at which the load is applied can be expressed as
u = CP
(2.12)
P
u
2a
Fig. 2.4 Component with a crack
where C is the compliance. The objective now is to find the energy release rate G in terms of
change of compliance with respect to the crack length a.
In order to provide a feel to the reader, we have chosen the case of a double cantilever beam
(DCB), which is made by splitting a beam on one end (Fig. 2.5). However, the proof is general and
Energy Release Rate
17
the results can be applied to other problems. Further, we have chosen the case in which the crack
is at mid-plane of the beam, with both cantilevers having identical geometry.
This problem is solved for two extreme cases: (i) constant load P in which the displacement of
load point increases as the crack grows and (ii) constant displacement where load decreases with
crack growth.
u
a
a
P
Da
P
(a)
(b)
Fig. 2.5 DCB specimen loaded with a constant load
Constant Load: In the case of a constant load (Fig. 2.5), the cantilever ends move away from
the crack plane with the advancing crack and thus work Pu is being done on the specimen.
Consequently, the cantilevers are flexed more and they absorb a part of energy out of the external
work. The remaining energy is utilized to extend the crack by Da. For this system strain energy U
and work done Wext are
1
Pu
2
Wext = Pu
The potential energy becomes
U=
1
1
Pu – Pu = –
Pu
(2.13)
2
2
Substituting the expression of P in Eq. (2.10) and making use of the fact that P remains constant,
we obtain:
P = U – Wext =
P du
2B da
Substituting u from Eq. (2.12), we obtain:
G=
P 2 dC
(2.14)
2B da
The equation provides another way to determine the energy release rate—just by finding the rate
of change of compliance. Also, the simple expression of this equation justifies our choice of
displacement-load equation in terms of compliance rather than in stiffness.
G=
18
Elements of Fracture Mechanics
Fixed Grip: The analysis for the case of fixed grip is quite different; but the result would
indeed be the same. As the crack advances, no external work is done on the system because the
external load is not allowed to move (Fig. 2.6). What is the mechanism of energy release required
for the crack growth? With the increase in the crack length, the cantilevers of Fig. 2.6a are relaxed
to acquire a smaller curvature (Fig. 2.6b). The configuration of Fig. 2.6b has less strain energy
because the energy stored in a cantilever is proportional to the square of the curvature. The body,
therefore, continuously releases its strain energy with increasing crack length and if the release is
large enough to meet the demands of producing two new surfaces, the crack may grow. The
potential energy of the system is:
1
P = Pu
2
a
a
(a)
Da
(b)
Fig. 2.6 DCB specimen with fixed grips
Substituting in Eq. (2.10), we obtain:
G=–
u dP
2B da
Substituting P from Eq. (2.12) and realizing u remains constant, we obtain:
G=
u2 dC
2BC 2 da
which, on eliminating u using Eq. (2.12), provides:
G=
P 2 dC
2B da
It is worth noting that the result is same as that of the constant load case.
We deliberately solved the two cases separately to make the reader aware that energy to
propagate a crack may come from different sources. In the case of a constant load, energy
requirements of the crack surfaces are met by the external work done on the body. In fact, the
external work done is split into two parts, first part [50%, Eq. (2.13)] increases the strain energy,
as the cantilever deforms to higher curvature and the second part is released for the crack growth.
In the case of a fixed grip, the entire energy needed for the advancement of the crack is met by the
decrease in existing strain energy.
In practical applications, the cases are mixed. If P and u are changing simultaneously, as shown
in Fig. 2.7, it can be argued that, for a short while, P is changed by DP at constant u and, then, u is
Energy Release Rate
19
changed by D u with constant P. Such successive changes approximately follow the actual loading
path. Since there is no constraint on the magnitude of D u and DP, they can be made as small as we
like; so much so that the actual load-displacement curve is approached exactly. Since the expression
for finding G is same for either case, Eq. (2.14) is valid for any general kind of loading.
DP
Du
P
u
Fig. 2.7 General case
The parameter G is often chosen by many researchers to determine interlaminar toughness of
a composite laminate. A DCB specimen is pulled in a tensile machine at a slow cross-head speed
(Fig. 2.8a). A stage is reached (state A in the load-displacement curve of Fig. 2.8b) when the crack
starts growing. It is allowed to grow by a small distance (ª 5 to 10 mm) to state M, and the
machine is stopped and unloaded. From A to M, load decreases due to reduced stiffness, and
distance u increases because the machine is continuously pulling the cantilever all along. The
slope of the unloading curve MO is smaller than that of the loading curve OA. With this data in
our hand, we could now evaluate the energy release rate. In fact, we will prove that the area
OAM is the energy released when the crack moves through distance D a.
P
A
P
u
C
B
M
O
(a)
B¢ C¢
u
(b)
Fig. 2.8 DCB specimen of a polymer composite laminate
Line AM is split into horizontal and vertical segments as shown. From A to B, the crack advances
with constant displacement, and, therefore, the energy equivalent to the area of DOAB is released.
20
Elements of Fracture Mechanics
Along BC the crack propagates at constant load, adding external work given by the rectangle
BCC’B’. Half of it is used to increase the strain energy of the cantilever given by the area of DOBC.
The rest is available as energy release for the growth of the crack, which is also equal to the area
of DOBC. Thus, by moving from point A to C, the energy release is equivalent to the sum of the
areas of triangles OAB and OBC. The remaining segments along AM, with the same argument,
provide the energy release as we move towards M. These segments can be made as small as
possible, giving area OAM as the overall energy release. Thus,
GBDa = Area OAM
yielding,
G=
Area OAM
BD a
2.6.4 Change in the Strain Energy Approach
Another method, based only on the change in the strain energy, is also found convenient for
linear elastic materials to determine energy release rate. For constant displacement (fixed grip)
case, Eq. (2.8b) is simplified to:
dU
(2.15)
dA
This equation states that the decrease in the existing strain energy per unit area extension of crack
is the energy release rate. In case of a constant load case, we have already seen that:
G=–
dWext
dU
=2
(2.16)
dA
dA
This, in fact, is a general theorem for linear elastic materials and is known as Clapeyron’s
Theorem. Substituting Eq. (2.16) in Eq. (2.8b), we have,
dU
(2.17)
dA
We thus obtain the same form for G in both cases, except with a difference of the negative sign.
In the fixed grip case, the already existing strain energy decreases as the crack advances, whereas
in the constant load case, strain energy of the component increases, which is equal to half of he
external work. However, the rate of change of internal energy does not give G in the mixed case
with simultaneous variation of P and u. In such cases, the method of finding the rate of change of
compliance [Eq. (2.14)] is preferred.
Some professionals like to address energy release rate as strain energy release rate (SERR).
There is nothing wrong in calling a phenomenon or object by any name as long as there is no
confusion in communication. Rigorously speaking, strain energy is not released in a constant
load case; only a part of the work done is released. We will, therefore, not be using SERR for
energy release rate in this book.
G=
Example 2.1 Determine the energy release rate for an edge crack loaded as shown in Fig. 2.9.
Solution: This is a case of constant load and, therefore, we can determine G using Eq. (2.17). For
each cantilever,
Energy Release Rate
21
B
h
M
h
a
M
Fig. 2.9 Edge-cracked body of Example 2.1
U=
1
Ú 2 Mk da
where k is the curvature of the beam and obtained as:
k=
M
12M
M
=
=
3
EI
EhB3
E( hB /12)
where E is the modulus and I the moment of inertia of the beam’s cross-section. Accounting for
strain energy in both the cantilevers,
a
1 Ê 12 M ˆ
12M 2 a
da
=
MÁ
˜
2 Ë EhB 3 ¯
EhB 3
0
U = 2Ú
Invoking Eq. (2.17), we have
G=
1 Ê dU ˆ
12M 2
ÁË
˜¯ =
B da
EhB4
2.7 ENERGY R ELEASE R ATE OF DCB SPECIMEN
The energy release rate can be obtained if the variation of compliance with respect to crack length
is known. For some cases, a relation between the applied force P and the displacement u can be
obtained by using field equations of solid mechanics. Then, by differentiating the ratio of u/P
and substituting in Eq. (2.14), the energy release rate is determined. In this section, we would
apply the method to a DCB specimen.
The deflection of a cantilever beam, d, caused by load P applied at the free end is well-known
and is given as,
d=
1 P l3
3E I
22
Elements of Fracture Mechanics
where l, E and I are the length, modulus, and moment of inertia of the beam’s cross-section,
respectively. For a DCB specimen with one cantilever-end attached to a fixed jaw of a tensile
machine, the deflection of the moving jaw is twice the deflection of one cantilever and, therefore,
for crack length a, the displacement u becomes
u=
2 P a3
3E I
(2.18)
leading to,
2 a3
u
=
(2.19)
P
3EI
For the rectangular cross-section of cantilevers, I = Bh3/12, where h is the depth of a cantilever
and B is the thickness of the DCB specimen. Substituting I in Eq. (2.19) we obtain
C=
a3
EBh3
Differentiating and substituting in Eq. (2.14), we obtain
C=8
12 a2 P 2
(2.20)
E B2 h3
The depth h plays a dominant role because deflection of the cantilever depends prominently
upon its depth. A cantilever of high depth flexes only by a small amount and therefore has a poor
capability of storing strain energy. It is to be noted here that energy release rate depends on the
capacity of the body to store strain energy. When the crack extends with fixed grip condition, the
energy release comes from the decrease in strain energy and if the capability to store energy is
small, the energy release rate is also small. In case of a constant load, the release of energy comes
from the external work, but it is equal to increase in strain energy. Therefore, a body with low
capability of storing strain energy provides small values of energy release rate.
The thickness B also controls the deflection of the cantilever and therefore a beam of larger
thickness would make the beam less flexible and provide a smaller energy release rate. Similarly,
material property, modulus E, also governs the deflection; a stiff material like steel does not allow
large deflection and, therefore, releases less energy in comparison to low modulus materials like
glass fiber composites or aluminum.
From Eq. (2.20), we should make one more observation which is disturbing from the designers’
point of view. If load P is maintained constant, energy release rate keeps on increasing with the
square of the crack length, whereas the surface energy required per unit surface area does not
depend much on the crack length for most engineering applications. Because of the excess energy,
which increases rapidly with the crack length, the crack is likely to attain high velocity causing
catastrophic failures. Designers find it very difficult to devise effective and practical methods to
arrest cracks that have become unstable. Therefore, while designing a component, an all-out effort
is made not to allow a crack to become unstable. In critical components, like an axle of a locomotive
wagon, the entire body is scanned with modern techniques (ultrasonic flaw detector, etc.) to make
sure that it does not contain a crack longer than the predetermined safe length.
GI =
Energy Release Rate
23
Example 2.2 Determine the critical energy release rate of a DCB specimen loaded in a tensile
testing machine. The thickness of the DCB specimen is 30 mm, depth of each cantilever 12 mm
and crack length 50 mm. It is made of a hardened steel with the modulus of 207 GPa and the crack
is about to propagate at 15405 N pulling load.
Solution Invoking Eq. (2.20), we have
GIc =
12 ¥ (0.050 m)2 ¥ (15405 N)2
12 a2 P 2
=
(207 ¥ 109 Pa) ¥ (0.030 m)2 ¥ (0.012 m)3
E B2 h3
= 22.1 kJ/m2
Example 2.3 Determine the shape of the DCB specimen if GI is to remain constant with the
growth of the crack. The specimen is loaded in the constant load mode. Determine the depth h of
the specimen beyond the crack tip if thickness of the specimen remains constant (B = 30 mm). The
initial crack length is 50 mm, modulus 207 GPa and depth of each cantilever 12 mm up to the
initial crack length.
Solution: For GI to remain constant in Eq. (2.20)
GI =
12a2 P 2
= constant
EB 2 h 3
leading to,
EB2 GI
a2
=
= constant
3
h
12P 2
But the constant is determined with initial conditions as
constant =
a02
(0.050 m)2
=
h03
(0.012 m)3
= 1447 m–1
yielding to,
h=
a2/3
1/3
(1447)
= 0.0884a2/3
P
2h
P
a
Fig. 2.10 DCB specimen with constant GI
For a = 0.08 m, the above relation gives h = 0.0164 m (=16.4 mm). The resulting shape of the
specimen is shown in Fig. 2.10. Note that the cantilever depth increases with the growing crack;
the cantilever with larger depth releases smaller energy because its capacity to store energy is
smaller, thus compensating the effect of increased crack length.
24
Elements of Fracture Mechanics
2.8 ANELASTIC D EFORMATION
AT
C RACK-TIP
To advance a crack, Griffith correctly identified that energy is required which is consumed in forming
two new surfaces. As discussed earlier in this chapter, the surface energy of solids is quite small and
only cracks in brittle materials (e.g., diamonds, window glass panes) advance by such criterion. For
most of the engineering materials (metals, polymer, etc.), a lot more energy is required in addition
to the surface energy in order to grow a crack. Therefore, besides surface energy of the solids, some
other mechanisms are also operative, which absorb large amounts of energy.
Inglis [2.10] showed that the stresses at crack-tips are quite large; so large that they cause an
anelastic deformation in front of the crack-tip. The anelastic deformation, such as plastic flow in
metals, is mostly irreversible and if the stresses are released the body will not attain its original
configuration near the crack-tip. The energy that causes this anelastic behavior is eventually
converted into heat energy and is lost to the surroundings. Thus, the anelastic deformation
dissipates work energy into heat energy.
In metals, the plastic deformation in the vicinity of the crack-tip is caused mainly by motion
and generation of dislocations, rotation of grains and grain boundaries, formation of voids, etc. If
the material is of low yield stress, the size of the plastic zone is large.
The plastic zone is very useful to designers who like to avoid fracture failure in machine
components. A large plastic zone means that a large amount of energy is required to advance a
crack-tip because a newer portion of the component is being continuously subjected to plastic
deformation. The energy is thus continuously being dissipated by the advancing crack tip to the
plastic deformation. This is analogous to the preparation of an agricultural field with a plough
which goes on throwing soil on either side. If the plough is dug deeper, more soil is upturned and
this requires more energy. On the contrary, if the plough is not inserted into the ground at all,
hardly any energy is required. The plastic deformation is thus a boon in disguise. A large plastic
zone introduces plastic deformation to greater depth in the cracked surfaces as shown in
Fig. 2.11. To understand the concepts, one may even express overall surface energy g as sum of
natural surface energy g n which a surface possesses even if it has not been subjected to any plastic
deformation and the surface energy g p which is caused by the plastic deformation near the cracked
surfaces. The simple relation is written as,
g = gn + g p
(2.21)
Depth of plastic
deformation
Plastic zone
Crack growth
Fig. 2.11 Increase of the effective surface energy due to plastic deformation
Energy Release Rate
25
For most structural metals used in our daily life, gp is several orders higher than gn and, therefore,
gn may as well be ignored (Table 2.2). However, Eq. (2.21) is not a convenient way to formulate
problems in fracture mechanics because there is always an uncertainty in the magnitude of gp.
TABLE 2.2 Representative values of gn and gp for some common materials
g n (J/m2)
g p (J/m 2)
Mild Steel
1.20
125,000
Alloy Steel
Aluminum Alloy
1.20
0.60
15,000
4,000
Material
In case of non-metals such as polymers, anelastic deformation near the crack tip is governed
by mechanisms different from those in metals. For example, polymer chains align themselves
parallely to each other under high stress and dissipate energy. The energy dissipation is several
orders more than the natural surface energy.
To sum up, the anelastic deformation in front of a crack-tip demands a large energy release
rate. Therefore, for most of the cracks in a body, the energy release rate is not high enough to
make them critical. Consequently, the cracks remain sleeping or dormant.
2.9 CRACK RESISTANCE
For a crack to grow, the crack resistance (R) is the energy required by the crack per unit increase
in area. It characterizes the material behavior. For most engineering materials, crack resistance
increases with the crack length as shown in Fig. 2.12. A minimum value Ri is needed to make the
crack grow. Crack resistance depends on the plastic zone size. In a crack with a large plastic zone
size, high energy is required to grow the crack as more material is subjected to plastic deformation.
A substantial portion of the energy is lost eventually to the surroundings.
R
Ri
a
Fig. 2.12 R-curve of ductile materials
Why does R increase with a? As the crack advances, plastic zone size becomes larger which, in
turn, requires higher energy for crack growth. Interestingly, the shape of the R-curve is not found
26
Elements of Fracture Mechanics
to depend much on the initial length of a crack for most of the engineering materials. However,
R-curve depends considerably on the temperature and the thickness of a plate. The dependence
on temperature is understandable because many other properties (yield stress, etc.) depend on
temperature. Dependence of the R-curve on a plate’s thickness involves a geometrical parameter
and makes it difficult to define toughness as an intrinsic property of a material. This aspect will
be taken up again in this chapter.
2.10 STABLE AND UNSTABLE CRACK GROWTH
We have already realized that for a crack to grow the energy release rate of a crack must be
greater than the crack resistance. We have now developed an adequate background to explore
other requirements for a crack to become unstable.
Consider a large plate with a centre crack length 2a, loaded in Mode I by stress s (Fig. 2.13). We
would explore how much energy is released at a constant stress s1 if the crack length is varied. It
will be shown in Chapter 4 that G = s 2p a/E for plane stress and G = (1 – v2) s 2p a/E for plane
strain cases. Note that G increases with crack length as shown in Fig. 2.13. If stress s1 is small and
the G-curve intersects the R-curve below Ri , the crack will not propagate because the energy
release rate is not adequate. The curve of higher stress s2 just passes through Ri and still the crack
does not advance because with Da increase in crack length, the energy release is smaller than the
requirement. If the stress is further increased to s3, G exceeds R for crack having length between
a0 and a3 and the crack is likely to grow. However, with the increase in crack length the difference
between G and R diminishes to zero at point B. There will not be any further advancement of the
crack. Therefore, the crack may advance from length a0 to length a3 only.
s
s4
G, R
R
s3
2a
B
s2
Ri
s1
s
O
a0 a3
a4
Fig. 2.13 (a) Centre-cracked plate, and (b) stable crack growth
If the applied stress increases gradually, point B shifts to the right (Fig. 2.13). In other words,
the crack grows slowly (stable crack growth). For stress s4 , G-curve just touches the R-curve
which means energy release rate is higher for all crack lengths except, of course, at the crack
length a4 where they are equal. Even a slightly higher stress or a perturbation makes the crack
Energy Release Rate
27
grow. As soon as the crack length increases, the difference between G and R grows, which provides
excess energy to the crack. As a result, the crack gains velocity, ending up in a catastrophic failure.
Thus, for a crack to grow and become critical, two conditions are necessary
G≥R
(2.21a)
dG
dR
≥
(2.21b)
da
da
The stable crack growth, at times, is useful in detecting a potential danger. Once a stable crack is
detected in a component, a remedial action should be taken either by finding ways to arrest the
crack or replacing the defective component with a new component.
Even if the above two conditions are fulfilled, the crack may not extend. The local conditions at
the crack-tip are quite important. If the tip is blunt, stresses at the crack-tip are not very high as it
is clear from the Inglis solution [Eq. (2.1)]. Then, the crack is unlikely to propagate, although the
energy criteria are met. An analogy may be drawn with some stone pieces placed right in front of
the tires of a car—the powerful engine finds itself helpless although the car is capable of moving
once the stone pieces are removed.
A designer should not feel assured in cases where cracks are of long lengths, satisfying
inequalities 2.21, but have blunt ends such as a hole at the crack-tip. Such cases are like sitting on
a volcanic mountain which may erupt at any time. On the surface of a blunt crack front, a small
length crack with a sharp tip may develop due to some unforeseen factors such as bad
workmanship, fatigue or environmental degradation. The effective crack length then is the sum
of the length of the blunt and the small crack. Under such conditions the crack may grow easily.
2.11 R-CURVE FOR BRITTLE CRACKS
In a brittle material, the size of the plastic zone in the vicinity of a crack tip is negligible and even
the growth of the plastic zone is negligible with the advancement of the crack. Consequently, the
R-curve rises vertically at the given crack length (Fig. 2.14) and then turns horizontal like a step
function.
R, G
G
R
a
Fig. 2.14 R-curve of brittle materials
As soon as the stress is large enough to make G overcome R, the crack advances. Note that
stable growth does not occur. The difference between G and R increases rapidly with increasing
crack length and the crack runs catastrophically.
28
Elements of Fracture Mechanics
2.12 THIN PLATE VS T HICK PLATE
From the point of view of fracture, which plate is tougher, thick or thin? Once I organized a
trekking trip in the Grand Canyon, USA. We went down, right up to the Colorado river; i.e., 5000
ft. down and then we climbed up 5000 ft. in a day covering a distance of nearly 45 km. One of my
team-mates, who, even after being a little overweight was fit enough to be admitted in the Indian
Army, had a very hard time climbing up the 5000 ft. On the other hand, there was another fellow,
extremely lean and thin, climbed without any problem and that too two hours ahead of the other
trekkers. From this incident the reader might have guessed by now that the thin plate is tougher;
but the question still remains, why?
It has already been argued that because of plastic deformation at the crack-tip, a large amount
0
of energy is dissipated
near the cracked surfaces. If the material yields easily and the plastic zone
size is large at the crack-tip, the crack will not grow at low values of the energy release rate. We
shall prove that in a thin plate (plane stress) the plastic zone size is considerably bigger than the
plastic zone size in a thick plate (plane strain). If this is so, a thin plate is tougher against crack
growth.
We will first consider yielding in the vicinity of the crack-tip in a thin plate. Both the free
surfaces of the plate are traction free and, therefore, the three stress components s33, s31 and s32
(Fig. 2.15a) are zero on the surfaces. Also, they are negligible small even at all interior points for
most of the engineering applications because the plate is thin. Note the s33 becomes one of the
three principal stresses and will hence forth be called s3.
t
tmax
x3
s
x1
x2
s3
s2
s1
s
s
(a)
(b)
Fig. 2.15 (a) A thin plate with a center crack and (b) principal stresses with a large
value of maximum shear stress
Other two principal stresses s1 and s2 lie in the plane of the plate. To avoid complex algebra
and emphasize more on the concepts, we explore the nature of stresses on points which are ahead
of the crack-tip and are close to the plane of the crack for Mode I cases. Shear stresses are small
near this plane due to the symmetry of the problem.
Normal stress s22 is large and tensile, contributing dominantly to the highest principal stress
s1. The other normal stress s11 is also tensile as the material is not free to move in the plane of the
plate. It contributes to next highest principal stress s2 as shown in Fig. 2.15b. For estimating the
Energy Release Rate
29
size of the plastic zone, we will now make use of Tresca Yield Criterion, which is based on
maximum shear stress. The maximum shear stress can easily be obtained by drawing all the three
Mohr circles. s1 being the largest is on the far right in Fig. 2.15b and s3 at the origin with s2 lying
in between. The radius of the largest Mohr circle, between s1 and s3, provides the maximum
shear stress tmax. It is worth noting that tmax is quite large in cases of thin plates resulting in the
generation of a plastic zone of a large size.
A thick plate with a Mode I crack, s2 and s1 are again tensile stresses as they are for the thin
plate. Due to tensile s1 and s2, the material tends to stretch in the plane of the plate, which in turn
tends to make the plate thinner. However, the plate is thick and constrained; the material is not
able to flow in x3 direction. As a result, tensile stress develops along x3 direction. Fig. 2.16 shows
all the three Mohr circles. For plane strain (thick plate), s3 is finite and only marginally smaller
than s1 making the largest Mohr circle substantially smaller than the largest Mohr circle of plane
stress. Thus, tmax is considerably lower in a thick plate. In fact, the material in front of the cracktip is locked and is not free to deform. In other words, thick plates do not allow the generation of
large plastic zone sizes. Hence, they are not as capable of dissipating energy as thin plates are.
These aspects will be discussed in detail in Chapter 5.
x3
s
t
tmax
x1
x2
s3
s2
s1
s
s
(a)
(b)
Fig. 2.16 (a) A thick plate with a center crack and (b) principal stresses with a
low value of maximum shear stress
When the Alaska pipeline to transport crude oil from the interiors of the Alaska state to the
sea-shore was being designed, there was a constant tussle between the conventional designers
and the modern experts of fracture mechanics. The conventional designers wanted the pipe line
to be made of thick wall so that stresses will be smaller and the material would not fail due to
yielding. The experts of fracture mechanics wanted to choose smaller wall thickness because then
the wall of the pipe would be loaded in the plane stress and be tough against crack growth.
2.13 CRITICAL E NERGY R ELEASE R ATE
R-curve depends on the thickness of a plate if it is loaded in a plane stress. Figure 2.17 shows
three R-curves for plane stress for three thicknesses, with the corresponding critical energy release
(2)
(3)
rates G(1)
c G c and G c .
30
Elements of Fracture Mechanics
B1
B2
B3
G (2)
c
G (3)
c
Plane strain
G, R
G (1)
c
Plane stress
B1 < B2 < B3
GIc
2a
B
a0
s
Crack length
Fig. 2.17 Critical energy release rate of plane strain and plate stress
When the thickness of the plate is increased further, it becomes a combined case of plane stress
and plane strain. Close to the free surfaces, material deforms in plane stress and, in the interior,
plane strain conditions exist. On thicker plates, the component of plane strain dominates; so much
so, that a stage is reached beyond which the effects of thickness are not felt on the R-curve as
shown in Fig. 2.17. Then the energy release rate required to cause the unstable growth no longer
depends on plate thickness. This is called critical energy release rate and is denoted by GIc where
‘I’ stands for Mode I. The critical energy release rate becomes the property of a material. Table 2.3
provides representative values of plane strain GIc of some widely used engineering materials.
Design engineers like to use components made of thin cross-sections so that they are loaded in
plane stress. For example, the web and the flange of an I-beam should be made of small thicknesses.
An irony is that, when designers refer to literatures or handbooks, the critical energy release rate
is available only for plane strain and not for plane stress. One may argue that by using the critical
energy release rate of plane strain we have conservative design. But, the resulting design may
sometimes turn out to be over-designed and non-feasible. For example, the structure of an airplane
is designed with a low factor of safety for restricting its dead weight. In such cases, the critical
energy release rate is determined for the thickness of the sheet chosen by the designers.
Experimental techniques to determine the critical energy release rate are not developed well
except in some special cases like determining the interlaminar Gc of fiber composite laminates.
Details of experimental techniques are presented in Chapter 8.
The concept of the energy release rate is not conveniently applicable to materials with large
plastic zone at the vicinity of the crack tip. In the analysis discussed so far in this chapter, the
material is assumed to be linearly elastic and if the plastic zone is small, its effect may be neglected
in determining the energy release rate. However, with a large plastic zone at the crack tip, the
analysis should be done more carefully by accounting for the constitutive relations of plasticity
within the plastic zone. The parameter J-Integral deals with such problems more effectively (Ch. 6).
Energy Release Rate
31
TABLE 2.3 Representative plane strain GIc of some common materials
Material
Mild Steel
Alloy Steel (s *ys = 1070)
EN 24 (U.K.)
4340 (U.S.A.)
40Ni2Cr1Mo28 (I.S.)
Aluminum 7075-T6
Titanium Ti-6Al-4V
Perspex (PMMA)
PVC
GIc (J/m2)
ª250,000
30,000
8,000
29,000
800
4,500
* Yield Stress in MPa
2.14 CLOSURE
In this chapter, the problem of fracture has been discussed purely through energy approach
without worrying about existence of very large stresses in the vicinity of the crack tip. To sum up,
there should be an energy release from the system to make a crack move. The energy release is
measured by the parameter, energy release rate (G). Also, energy is required to form two new
cracked surfaces and is accounted by another parameter, crack resistance (R). The necessary
requirements for the crack advance are (i) G ≥ R and (ii) dG/da ≥ dR/da. The value of G, which
satisfies these two conditions, becomes the material property known as critical energy release
rate. For thin plates, the critical energy release rate depends on the thickness of a plate, whereas
it is independent of the thickness of thick plates.
QUESTIONS
1. Why is the surface of a solid associated with surface energy (or free energy)?
What is an approximate value of free energy of the surface of a metal?
2. Actual energy required in a ductile material to create two new surfaces through the crack
growth is several orders higher than the surface energy of solids. Why so?
3. Why does the compliance of a component increase with the growth of a crack?
4. Does the strain energy of a component increase with the crack growth?
5. Why does negative sign appear in the expression of G defined through Eq. (2.9)?
6. How does the rate of change of strain energy give G for constant load case or
constant deflection case? Is this approach valid for non-linear elastic materials?
7. Why is failure catastrophic once a crack starts growing in a DCB specimen?
8. What are different mechanisms of anelastic deformation in a polymer?
9. If the load is increased gradually on a component made of a ductile material, why do we
obtain stable crack growth before the catastrophic failure?
Elements of Fracture Mechanics
32
10. Why does a brittle material not have stable crack growth?
11. Why are thin plates tougher in comparison to thick plates?
12. Why is critical energy release rate not given in handbooks for thin plates?
PROBLEMS
1. Determine the energy release rate of a DCB specimen through change in strain energy
approach for constant load.
2. Determine the energy release rate, using elementary beam analysis, for the configurations
given in Fig. 2.18 (h<<a).
h
B
h
B
M
h
P
h
M
a
P
a
(a)
(b)
h
P
h
B
h
P
h
P
a
B
(c)
P
2a
(d)
Fig. 2.18 The figure of Problem 2
3. GI is determined for a DCB specimen with a load at the end of each cantilever in
Section 2.7 by neglecting the effect of the shear stress. If the strain energy of the shear
force is also considered, find GI.
Energy Release Rate
33
4. Consider a plate with a crack of length a and lateral load P acting on the corners as shown
in Fig. 2.19. Find the energy release rate.
[q = TL / Km where q is the angle of twist, T twisting moment, L length, m shear modulus
and
K=
bh 3 È 16
hÊ
h4 ˆ ˘
Í - 3.36 Á 1 ˙
16 ÎÍ 3
bË
12b 4 ˜¯ ˚˙
h
P
a
P
b
b
b =5
h
h<<b
<<a
Fig. 2.19 The figure of Problem 4
5. The load on a 30 mm thick plate with an edge crack of 50 mm length was increased very
slowly and the displacement of the load point was monitored. It was observed that at the
load of 2100 N and displacement u = 4.1 mm, the crack started growing. The rate of crack
growth was much faster than the rate at which the load increased and therefore the crack
essentially was grown at the load of 2100 N. Through an optical recording using a rapid
camera it was found that the crack grew up to 65 mm length with the rapid increase in
displacement to u = 7.5 mm. Determine the critical energy release rate.
6. A large plate of 36 mm thickness with an edge crack of a = 32 mm length is pulled very
slowly under displacement control loading. At the displacement of 7.2 mm, when the
recorded load is 2750 N, the crack starts growing. At a = 41.7 mm, the crack is arrested
and the load decreases to 1560 N. Determine the critical energy release rate.
7. A double cantilever beam of a fiber composite laminate with an initial pre-crack is loaded
under displacement control to the critical load and the crack is allowed to grow by a small
length. Then, the specimen is unloaded to zero load. Now, the specimen with a larger
crack length is loaded again to the new critical load and allowed to grow further by a
small length. Such load cycles are repeated five times and compliance is obtained during
each cycle. The data is summarized in Table 2.4. The load displacement curve remains
elastic up to the critical point and the thickness of the specimen is B = 30 mm. Determine
the critical strain energy rate for each segment of crack growth through the approach of
finding the area enclosed in the corresponding triangle of the load-displacement recording.
34
Elements of Fracture Mechanics
TABLE 2.4 The table of Problem 7
Crack length a (mm)
32
42
50
58
67
Compliance C (m/N)
33.5
68.9
121.0
191.0
294.0
¥
¥
¥
¥
¥
–6
10
10 – 6
10 – 6
10 – 6
10 – 6
Critical load P (N)
94.3
78.8
64.0
53.5
44.0
REFERENCES
2.1 Kanninen, M.F. and Popelar, C . H. (1985). Advanced Fracture Mechanics, Oxford University
Press, New York.
2.2 Gdoutos, E.E. (1993). Fracture Mechanics—An Introduction, Kluwer Academic Publishers,
Dordrecht.
2.3 Broek, D. (1982). Elementary Engineering Fracture Mechanics, Martinus Nijhoff Publishers,
The Hague.
2.4 Meguid, S.A. (1989). Engineering Fracture Mechanics, Elsevier Applied Science, London.
2.5 Barsom, J. M. and Rolfe, S.T. (1987). Fracture and Fatigue Control in Structures, PrenticeHall, Inc., Englewood Cliffs, New Jersey.
2.6 Hertzberg, R. W. (1989). Deformation and Fracture Mechanics of Engineering Materials, John
Wiley and Sons, New York.
2.7 Knott, J.F. (1973). Fundamentals of Fracture Mechanics, John Wiley and Sons, New York.
2.8 Griffith, A.A. (1921). The Phenomena of Rupture and Flow in Solids, Philosophical
Transactions of the Royal Society of London, A 221, pp 163–169.
2.9 Griffith, A.A. (1924). The Theory of Rupture, Proceedings of the First International Conference
of Applied Mechanics, Delft.
2.10 Inglis, C. E. (1913). Stress in a Plate due to the Presence of Cracks and Sharp Corners,
Transactions of the Institute of Naval Architects, 55, pp. 219–241.
2.11 Irwin, G. R. (1948). Fracture Dynamics: Fracturing of Metals, American Society for Metals,
Cleveland, pp. 147–166.
2.12 Orowan, E. (1948). Fracture and Strength of Solids, Reports on Progress in Physics, XII,
p. 185.
2.13 Anderson, T. L. (2004). Fracture Mechanics: Fundamentals and Applications, CRC, Press-Book.
2.14 Sanford, R. J. (2003). Principles of Fracture Mechanics, Prentice Hall, Upper Saddle River.
2.15 Janssen, M., Zuidema, J. & Wanhill, R.J.H. (2004). Fracture Mechanics, Spon Press, Abingdon.
2.16 Ramesh, K. (2007). e-Book on Engineering Fracture Mechanics, IIT Madras,
URL: http//apm.iitm.ac.in/smlab/kramesh/book_4_htm.
Chapter
3
Stress Intensity Factor
... nature is intelligent because it is too complicated to be called anything else.
Deepak Chopra
3.1 INTRODUCTION
3.1.1 Why Should Investigations be Closer to the Crack Tip?
Today, thinking big often implies focusing on minute details. Huge amounts of atomic energy is
generated, both for peaceful and destructive purposes, by manipulating atoms which are invisible
even under powerful microscopes. Similarly, in order to understand how crystalline materials
(e.g., metals) deform plastically, we should understand the behavior of dislocations, which are
imperfections and look like threads. Dislocations too are very small in diameter, but can be viewed
under a powerful microscope. A crack front exists within a material, like a line running from one
region of the body to another. The vicinity of a crack tip offers interesting information as the
magnitudes of stress components are extremely high. When we want to break a wooden stick by
bending it we make a notch. The notch creates high stresses, which in turn make the notch tip
move rather easily.
Knowing the stress or displacement field in the vicinity of a crack tip is very useful. A material
scientist may devise ways to develop new materials which can diffuse high stresses at the crack
tip. A designer may modify some features such as notches, cutouts, keyways, etc., to minimize
stresses. An experimentalist can think of methods of characterizing cracks by measuring stresses
or strains near the crack tip. One of the biggest advantages is that stress analysis leads to define a
parameter, stress intensity factor (SIF) to characterize a crack. In comparison to energy release
rate, SIF is simpler for a designer and easier for laboratory measurements, so as to determine
material properties.
3.1.2 Linear Elastic Fracture Mechanics (LEFM)
In a brittle material, the material remains elastic even at the crack tip where stresses are high.
Most engineering materials do not fall in this category. Diamonds are known to be quite brittle
36
Elements of Fracture Mechanics
(elastic) even in the vicinity of a crack tip. Other brittle materials like window panes are known to
have some anelastic deformation close to the crack tip. This chapter develops analysis of brittle
materials only; the analysis does not account for plastic deformation close to the crack tip. A
natural question arises—when most materials are not brittle, why should we make detailed
analysis of brittle materials? Obviously, analysis of brittle materials is far simpler than the analysis
of a material having a plastic zone at the crack tip. The presence of a plastic zone means that two
kinds of stress-strain behaviors should be incorporated, plastic behavior inside the plastic zone
and elastic behavior outside it. At the same time, it is difficult to evaluate the interface between
the two zones. Furthermore, the material behavior inside the plastic zone is complex to model
and in some materials, especially those which exhibit pronounced Bauschinger effect, the material
behavior is extremely difficult to account for. Therefore, for the analysis of elastic-plastic fracture
mechanics (EPFM), problems are usually solved through numerical analysis.
By confining our attention to elastic (or brittle) materials, we would be able to obtain closed
form solutions to many problems. Also, we will learn how to deal with the singularity (infinite
stresses at the crack tip) involved. There is another advantage in developing solutions of elastic
crack problems. In many real life cases where the plastic zone size is quite small in comparison to
the crack length, the contribution of the plastic zone in an elastic analysis may be neglected. That
is, if stress fields in such cases are determined for purely elastic and elastic-plastic cases separately,
the difference between the two is small enough to be considered negligible. A large number of
engineering problems of practical applications fall in this category and consequently elastic
analysis is good enough. This leads to an important subfield—linear elastic fracture mechanics
(LEFM), where only elastic analysis is carried out to determine stress and displacement fields
near a crack tip with characterizing parameters like the SIF. It is worth mentioning here that the
energy release rate (G) has also been formulated for LEFM.
3.2 STRESS
AND
DISPLACEMENT FIELDS IN ISOTROPIC ELASTIC MATERIALS
On what parameters does the stress field depend in the vicinity of the crack tip? It, of course,
depends on the external load, generally denoted by the far field stress s . The name, far field, may
mislead some readers; it refers to stress at points of the body where the influence of the high
stresses generated at the crack tip is negligible. The second important parameter, as discussed in
Chapter 2, is the crack length a. We also realize that the stress field varies near the tip significantly
from point to point and one must specify the coordinates at which stress components are to be
determined. Generally, polar coordinates (r, q ) are employed for describing the location of the
point. Further, the stress field depends upon geometry i.e., whether the crack is at the centre of
the body, at the edge, or at the off-centre. Under geometrical considerations, the effect of the
overall size of the specimen with respect to the crack length would also be included.
Let us express stress component s ij in the vicinity of crack tip as
s ij = f(s, a, r, q, geometry)
where i, j are the suffixes representing various stress components like s 11, s 22, s 12, s 33, etc. There
is a separate relation (function f ) for each stress component.
In engineering applications, most of the hardware components are in the form of plates.
Commonly used sections like angles, I-beams, channels, rectangular tubes, etc., are made of flat
plate sections. Even in a circular tube with a crack having its length much smaller than the diameter
Stress Intensity Factor
37
of the tube, the region around the crack tip may be regarded as a flat plate. Thus, a study of crack
in a flat plate encompasses many engineering applications.
In most problems of this book, x1-axis is chosen along the crack length, x3-axis along the normal
of the plate and x2-axis in the plane of the plate and normal to the crack length [Fig. 3.1(a)]. Also,
we have chosen to denote stress components with two suffixes because that is a better way of
representing the stress tensor. Consequently, we would be using symbol s for shear stress
components also; the suffixes will only differentiate shear components from normal components.
Further, we prefer to use suffixes 1, 2 and 3 in place of x, y and z, as used in many books.
Fig. 3.1
Axes with respect to the crack in a plate
Solving equations of solid mechanics to obtain expressions for s ij in the vicinity of the crack tip
is mathematically involved and, therefore, only the results would be presented and discussed in
this section. The derivations would be presented in Sections 3.4 and 3.5.
To present results, we have chosen the common case of a flat plate with a crack of length 2a and
far field stress s as shown in Fig. 3.2(a). The stress field at a general point H near the crack tip for
isotropic and linear elastic material in the flat plate for this Mode I case is
Fig. 3.2
(a) Infinite plate with a crack of length 2a subjected to a far field stress s, and
(b) definition of stress components at point H
38
Elements of Fracture Mechanics
s 11 =
s 22 =
s 12 =
s (p a)1/2
(2p r)1/2
s (p a)1/2
(2p r)1/2
s (p a)1/2
1/2
(2p r)
cos
qÈ
q
3q ˘
1 - sin sin
2 ÍÎ
2
2 ˙˚
(3.1a)
cos
qÈ
q
3q ˘
1 + sin sin
2 ÍÎ
2
2 ˙˚
(3.1b)
sin
q
q
3q
cos
cos
2
2
2
(3.1c)
For a thin plate, other stress components are negligible. In case of a thick plate s 33 = n (s 11 + s 22)
where n is the Poisson's Ratio of the material; the other two stress components (s 13, s 23) are
negligible.
It is clear from these equations that each stress component is proportional to the far field stress
s . The crack length appears under a square root and, therefore, its influence on stress components
is also prominent, but not to the extent of s .
The distance (r) between the crack tip and the point plays the most important role. It sits in the
denominator under a square root sign. If r becomes very small, the stress components, specially
s22 goes up steeply, so much so that for r Æ 0, s22 tends to be infinite. Such solutions are called
singular. In this case, it is known as square root singularity. In some special cases of fracture
mechanics, other kinds of singularities are encountered.
Displacement field for a plane strain near the crack tip for Mode I of Fig. 3.2(a) is given by
u1 =
s (p a)1/2 Ê r ˆ 1/2
qÈ
q˘
cos Í1 - 2n + sin 2 ˙
ÁË
˜
m
2Î
2˚
2p ¯
(3.2a)
u2 =
s (p a)1/2 Ê r ˆ 1/2
qÈ
q˘
sin Í 2 - 2n + cos2 ˙
ÁË
˜¯
m
2Î
2˚
2p
(3.2b)
u3 = 0
(3.2c)
where m is the shear modulus. The above equations do not contain any singularity, because
displacement is finite near the crack tip. Mathematically, displacement components are obtained
from stress equations by converting stress components to strain components and then integrating
the resulting expressions. During integration, the square root singularity disappears and
displacement components turn out to be proportional to the square root of the distance r. However,
one should note here that the Eqs (3.2a–c) are valid only in the close vicinity of the crack tip.
3.3 STRESS INTENSITY FACTOR
In the engineering field, a problem with two variables is much more difficult to solve than a
problem of one variable. A question may be raised here—can two independent variables be
combined to form a new independent variable? If the answer is yes, the solution to the problem is
likely to become much simpler. In the case of a wave propagation, solution even along one
dimension becomes complex because two variables, space (x) and time (t), are involved. In such
wave problems, we usually combine two variables to form a new variable, (x – ct), where c is the
Stress Intensity Factor
39
velocity of the sound. There are many other similar cases such as linear kinetic energy combining
two variables to one, linear momentum combining two variables to one, Reynolds Number in
fluid mechanics combining four independent variables to one, and Sommerfield Number in journal
bearing combining five variables into one.
If we look carefully into Eqs (3.1) and (3.2), we find that for a given geometry there are two
main variables, the far field stress s and the crack length a. Furthermore, in all the equations of
stress and displacement, s and a coexist as s a . Can this product be called by a different variable?
Now with several decades of research work, we find that it is advantageous to do so. This credit
goes to Irwin [3.1], who defined the new variable, stress intensity factor, and used the symbol K
after the name of his collaborator Kies [3.2]. He defined K as
KI = s (p a)1/2
(3.3)
There is no reason to have p in the above definition. It was included in the expression because
of some historical reasons which shall be explained in the next chapter. However, the stress
intensity factor KI is formally defined as
KI = (2pr)1/2 s 22 (r, q = 0) as r Æ 0
(3.4)
This definition can be checked easily by substituting Eq. (3.1b) in the formal definition [Eq. (3.4)],
the resulting expression will be the same as of Eq. (3.3).
We realize that, for stress or displacement fields, magnitude of s or a is immaterial as long as
s (p a)1/2 is same. This means a small crack length in a plate with high far field stress is equivalent
to a large crack length with small far field stress, provided K remains same. This combination of
s and a to form a new variable is regarded as a breakthrough in the field of fracture mechanics.
For Modes I–III, the stress intensity factor is written as KI , KII and KIII respectively, with subscript
in Roman numbers.
The stress and displacement Eqs (3.1) and (3.2) may now be written in terms of the stress
intensity factor. For Mode I problems of plane strain, they become
s 11 =
s 22 =
s 12 =
u1 =
1/2
cos
qÈ
q
3q ˘
1 – sin sin
Í
2Î
2
2 ˙˚
(3.5a)
1/2
cos
qÈ
q
3q ˘
1 + sin sin
2 ÍÎ
2
2 ˙˚
(3.5b)
KI
(2p r)
KI
(2p r)
q
q
KI
3q
sin cos cos
1/2
2
2
2
(2p r)
KI Ê r ˆ
Á
˜
m Ë 2p ¯
1/2
KI Ê r ˆ
Á
˜
m Ë 2p ¯
1/2
cos
qÈ
q˘
1 – 2n + sin 2 ˙
2 ÍÎ
2˚
(3.5c)
(3.5d)
q È
q˘
(3.5e)
2 – 2n + cos 2 ˙ .
Í
2Î
2˚
The stress intensity factor elegantly characterizes a crack, similar to energy release rate, G,
developed in Chapter 2. Equations (3.5a–e) need to be modified for bodies which are of finite
dimensions or where the crack tip is close to one of the free edges of the component. Closed form
expression for stress intensity factor is available only for simple cases and, therefore, determining
u2 =
sin
40
Elements of Fracture Mechanics
the stress intensity factor becomes a challenge for many practical cases. These days, numerical
techniques are widely used for this very purpose. However, for a body with a crack and known
boundary conditions, once the stress intensity factor is determined, the crack is characterized for a
designer and then, he can predict whether a crack in the work-component is likely to grow or not.
Stress and displacement equations for the center-cracked body are similar for other modes.
For Mode II in plane strain and far field stress s 12 = t (Fig. 3.3) with KII = t p a ,
Fig. 3.3 A centre-crack in an infinite plate loaded in Mode II
we have
s11 = –
s22 =
s12 =
K II
1/2
(2p r)
KII
1/2
(2p r)
KII
1/2
(2p r)
sin
qÈ
q
3q ˘
2 + cos cos
Í
2Î
2
2 ˙˚
sin
q
q
3q
cos cos
2
2
2
cos
qÈ
q
3q ˘
1 - sin sin
2 ÍÎ
2
2 ˙˚
KII
m
Ê r ˆ
ÁË ˜¯
2p
1/2
u1 =
KII
m
Ê r ˆ
ÁË ˜¯
2p
1/2
u2 =
sin
q
2
È
2q˘
ÍÎ 2 - 2n + cos 2 ˙˚
cos
q
2
È
2q˘
ÍÎ - 1 + 2n + sin 2 ˙˚
u3 = 0.
For Mode III and far field stress s23 = t (Fig. 3.4) with KIII = t p a , we have
s11 = s 22 = s 33 = s 12 = 0
Stress Intensity Factor
s13 = s23 =
KIII
(2p r )
1/2
KIII
(2p r )
1/2
sin
cos
41
q
2
q
2
u1 = u2 = 0
u3 =
Fig. 3.4
3.4 BACKGROUND
KIII Ê 2r ˆ
Á ˜
m Ëp¯
1/2
sin
q
2
A centre-crack in an infinite plate loaded in Mode III
FOR
MATHEMATICAL ANALYSIS
In the previous section, stress and displacement fields in the vicinity of a crack tip were presented
without providing proofs. This section would develop the mathematical base and the following
section would solve problems of all the three modes for simple geometries.
Although the field equations of solid mechanics are quite well-known, they would be presented
for plane problems for the sake of completeness. The solution to fracture problems would be
obtained by solving field equations of solid mechanics. Of course, proper boundary conditions
would be incorporated, including zero traction conditions (free surface) of the cracked faces. To a
beginner, some parts of this section may look strange and may be a bit difficult in the first reading
because such analysis is not usually seen in other fields of mechanics of solids. A straightforward
and simple analysis is not possible because the problem deals with a singularity. We would proceed
slowly as if we are trying to reach the centre of a flower bud by taking one petal out gently at a
time. This, of course, comes at a risk of annoying some of the readers who are already conversant
with this field.
42
Elements of Fracture Mechanics
3.4.1 Field Equations
There are three kinds of field equations [3.3] which are solved for a set of given boundary
conditions of a component. They are: (i) equilibrium equations relating to stress components, (ii)
strain-displacement relations and (iii) stress-strain relations for the given material of the
component. In case the problem is solved with stress or strain components as dependent variables,
an additional set of equations must be satisfied to ensure that during the deformation, a continuous
body remains continuous. This condition is called compatibility [3.4]. We will develop it because
most problems would be solved in stress components. In fact, the compatibility condition provides
the governing differential equation for many problems.
Equilibrium Equations: Equilibrium equations within a body are developed in any textbook of
solid mechanics and, therefore, they will not be derived here. For problems dealing with plates,
variation of stress components in the thickness direction is assumed to be negligible. For most
fracture mechanics problems, body force is not important and, therefore, we are left with the
following two differential equations of equilibrium:
∂s 11
∂ s 12
+
=0
∂x1
∂x 2
(3.6a)
∂s 12
∂ s 22
+
=0
∂x1
∂x2
(3.6b)
Strain-Displacement and Compatibility Relations: Only three strain-displacement relations are
involved in most problems of fracture mechanics dealing with plates. If u1 and u2 are displacement
components, strain components e11, e22 and e12 are expressed by well known relations, as:
e11 =
∂u1
∂x1
(3.7a)
e22 =
∂u2
∂x2
(3.7b)
1 È ∂u2 ∂u1 ˘
(3.7c)
+
Í
˙
2 Î ∂x1 ∂x2 ˚
It is to be noted here that the tensorial strain has been chosen, whose normal components are the
same as the normal components of the engineering strain, but a shear component is half the
corresponding shear component of the engineering strain. The compatibility conditions (necessary
relation between strain components) are obtained by eliminating u1 and u2 from the above
equations. Differentiating the equations, we have
e12 =
∂ 2 e 11
∂3 u1
=
∂x22
∂x1∂x22
∂ 2 e 22
∂ 3 u2
=
∂x12
∂x12 ∂x 2
∂ 2e 12
∂3 u1 ˘
1 È ∂3 u
= Í 2 2 +
˙
∂x1∂x2
2 Î ∂x1 ∂x2 ∂x1∂x22 ˚
Stress Intensity Factor
43
By substituting the first two equations in the last equation, and rearranging the terms, we obtain
∂ 2e 11
∂x22
+
∂ 2e 22
∂x12
-2
∂ 2e 12
=0
∂x1∂x2
(3.8)
This relation between the strain components ensures compatibility.
Stress-Strain Relations: For the linear isotropic materials deforming elastically, the stress-strain
relations are well known as
e11 =
1
Ès 11 – n (s 22 + s 33 ) ˘˚
EÎ
(3.9a)
e22 =
1
Ès 22 – n (s 11 + s 33 ) ˘˚
EÎ
(3.9b)
e33 =
1
Ès 33 – n (s 11 + s 22 ) ˘˚
EÎ
(3.9c)
e12 =
(1 + n )
s 12
=
s 12
2m
E
(3.9d)
where E is the Young's Modulus, m is the shear modulus and n is the Poisson's Ratio.
Plane Deformation: Consider a thin plate that is deformed in plane stress. On the free surfaces,
the out of plane stresses are zero and they are usually negligible in the interior points of the plate.
We thus assume
s13 = s23 = s33 = 0
Therefore, the plate carries only in-plane stresses. The stress-strain relations are simplified to
e11 =
1
Ès 11 – n s 22 ˚˘
EÎ
(3.10a)
e22 =
1
Ès 22 – n s 11 ˘˚
EÎ
(3.10b)
e12 =
(1 + n )
s 12
=
s12.
E
2m
(3.10c)
On the other hand, the plane strain case corresponds to a sufficiently thick plate for which (i)
Ê ∂
ˆ
= 0˜ .
displacement in x3 direction is restricted (u3 = 0) and (ii) variation in x3 direction is zero Á
∂
x
Ë 3
¯
These two conditions yield
e13 = e23 = e33 = 0
Therefore, for plane strain cases, simplified stress-strain equations are obtained by setting
e33 = 0 in Eq. (3.9c) to have
s33 = n (s11 + s22)
44
Elements of Fracture Mechanics
Substituting s33 in Eqs (3.9a) and (3.9b) and rearranging the terms, we have
e11 =
˘
(1 - n 2 ) È
n
s 22 ˙
Ís 11 E Î
1 -n
˚
(3.11a)
e22 =
˘
(1 - n 2 ) È
n
s 11 ˙
Ís 22 E Î
1 -n
˚
(3.11b)
Also, Eq. (3.9d) is manipulated to
e12 =
(1 - n ) (1 - n )
1 -n 2
s12 =
E (1 - n )
E
Ê
n ˆ
ÁË 1 + 1 - n ˜¯ s12
(3.11c)
In the above equations, we define
E¢ =
n¢=
E
1 -n 2
n
1 -n
(3.11d)
(3.11e)
Then, stress-strain relations for the plane strain become:
1
Ès 11 – n ¢ s 22 ˘˚
E¢ Î
1
e22 =
Ès 22 – n ¢ s 11 ˘˚
E¢ Î
e11 =
(3.12)
1 +n ¢
s12.
E¢
We thus note that the form of these equations is exactly the same as that of the corresponding
equations for plane stress cases [Eqs (3.10)].
e12 =
Biharmonic Differential Equation: When we glance at the field equations, we find that they are
too many of them to compute. Therefore, we shall reduce the number of equations by going to the
higher order of differential equations.
The analysis of determining stress field in the vicinity of a crack tip can be done either in stress,
strain or displacement components. Based on experience we find that it is convenient to solve the
differential equations with stress components as dependent variables for many problems of
fracture mechanics. This is because the boundary conditions, especially at the cracked faces, are
usually known in stress components. In fact, the cracked surfaces are generally traction free,
thereby making several stress components zero (e.g., s22 = s12 = 0).
Since we are planning to develop the differential equations in terms of stress components,
compatibility conditions should be invoked. Thus, the compatibility condition of Eq. (3.8), which
relates strain components should be changed into a relation between stress components. This is
done by substituting stress-strain relations [Eq. (3.10)] into the compatibility relation [Eq. (3.8)].
Thus, we have
∂2
∂2
∂2s 12
(s
–
ns
)
+
(s
–
ns
)
–
2(1
+
n)
=0
11
22
22
11
∂x1∂x2
∂x22
∂x12
(3.13)
Stress Intensity Factor
45
This equation has three dependent variables s11, s22, s12 and is still cumbersome to solve. To
make it manageable a new function F, known as Airy's Stress Function, is defined as:
s11 =
s22 =
∂2 F
∂x22
∂2 F
(3.14)
∂x12
s12 = –
∂2 F
∂x1∂x2
There is a definite purpose in coming to such a definition—they satisfy the equilibrium equations.
Therefore, we will not worry for the equilibrium conditions any more. Substituting Eq. (3.14) in
Eq. (3.13) and simplifying, we obtain the governing differential equation as
∂4 F
∂4F
2∂ 4 F
=0
+
+
∂x14 ∂x12 ∂x22 ∂x24
(3.15a)
This equation is generally written in more compact form by making use of the symbol
—2 =
Then, Eq. (3.15a) is expressed as
∂2
∂2
+ 2
2
∂x1 ∂x2
—2 (— 2F) = 0
or
—4 F = 0
(3.15b)
which is known as Biharmonic Equation.
If the above analysis is carried out for a plane strain [using Eq. (3.12) in place of Eq. (3.10)], the
governing differential equation turns out to be the same. This is because the stress-strain equations
of the plane stress and the plane strain have identical forms. The biharmonic solution may appear
to be deceptively simple in appearance, but the solution to even simple problems is often not
straightforward, because it involves fourth order partial derivatives.
3.4.2 Elementary Properties of Complex Variables
Stress and displacement fields in the vicinity of a crack tip have been obtained in several ways by
various investigators. Some like to solve it through complex variable techniques and others
without it. We will solve it through complex variables, as presented by Westergaard [3.5] in 1939.
Some of the readers may hesitate in the beginning to play with complex numbers but we feel that
it provides a short and simple route. It is like somebody travels from Delhi to Mumbai by an
airplane instead by a train; the airplane takes only one tenth of time, but is more complex.
It has been found convenient to form a complex variable z as
z = x1 + ix2
In fact, by combining two independent variables x1 and x2 into one complex variable z, we reduce
our difficulties of dealing with two independent variables in solving the biharmonic equation.
46
Elements of Fracture Mechanics
If F(z) is a complex function of variable z, it can be written as
F (z) = ReF1 + i ImF
In fact, F(z) forms a surface on the complex plane (Fig. 3.5). The function is analytic at a point z if
the derivative is the same in all directions [3.6]. This fact is quite useful and we equate derivative
of F(z) in x1 direction with its derivative in ix2 direction to obtain
F(z)
P
x1
x2
Fig. 3.5 Surface F(z) on a complex plane
F¢(z) =
∂F ∂ Re F
∂ Im F
=
+i
∂x1
∂x1
∂x1
F¢(z) =
∂F
∂ Re F ∂ Im F
= -i
+
∂ x2
∂ x2
i ∂ x2
Since F¢(z) = Re F¢ + i ImF¢, the equations are rewritten as
F¢(z) = ReF¢ + i ImF¢ =
∂ Re F
∂ Im F
+i
∂x1
∂x1
F¢(z) = ReF¢ + i ImF¢ = - i
∂ Re F ∂ Im F
+
∂x2
∂x2
Comparing real and imaginary parts, we obtain (Cauchy–Riemann relations):
∂ Re F
= ReF¢
∂x1
∂ Im F
= ImF¢
∂x1
(3.16)
∂ Re F
= – ImF¢
∂x2
∂ Im F
= ReF¢
∂ x2
All the four equations are useful for analysis because they provide the method for differentiating
the real and imaginary parts of F(z) with respect to x1 and x2.
Stress Intensity Factor
47
3.5 WESTERGAARD’S A PPROACH
One way to solve the biharmonic equation is to express F in terms of another complex function
Zl(z) for Mode I problems. In other words, Zl(z) is an intermediate solution which satisfies the
biharmonic equation. It is chosen appropriately by Westergaard to suit the general characteristics
of Mode I problems. Then to solve a specific problem, the form of Zl(z) is chosen to satisfy all the
boundary conditions of the problem. An analogy is drawn here with an airplane that goes from
Delhi to Bangalore. The intermediate solution ZI(z) is analogous to the airplane's flight which
does a major job of taking passengers from a far away distance to the city. Then, dispersing
passengers from the airport to various different parts of the city is analogous to solving various
problems of Mode I.
Similarly, there is a Westergaard intermediate solution ZII for the biharmonic equation for
Mode II problems. Since general characteristics of Mode II problems are different from those of
Mode I, ZII has a different form. In the analogy of air flights the airplane goes from Delhi to
Bangalore for Mode I problems whereas it flies from Delhi to Mumbai for Mode II problems.
The case of Mode III problems is usually simpler and can be managed without taking the help
of the biharmonic equation. The problem can be solved in terms of displacement component u3,
thus avoiding the compatibility equations. It would be shown that the governing differential
equation consists of only second order partial derivatives.
3.5.1 Mode I (Opening Mode)
For Mode I problems, F is expressed as
F = Re ZI + x2 Im ZI
(3.17)
where, ZI = dZI / dz and ZI = dZ / dz
We shall show that this expression of F satisfies the biharmonic equation. Also, components
s11, s22 and s12 would be expressed in terms of ZI(z) with the help of Eq. (3.14). What are the
advantages in constructing the Westergaard function ZI(z)? For many complex looking differential
equations, such as the biharmonic equation, there are no set ways of solving them. We keep trying
various solutions on the basis of feel, intuition or clairvoyance until one satisfies the differential
equation. Westergaard is credited here for offering a simple solution. The integrals of Z and Z
do not cause difficulties in determining the stress field of a problem, because ZI(z) is differentiated
at least twice and in the process, integrals are eliminated.
The proof of showing that the expression of F satisfies the biharmonic equation is straightforward, but it is carried out below for the sake of completeness. We would be adopting the
convention
dZI
dZI¢
dZI¢¢
Z¢I =
; Z¢¢I =
and Z¢¢¢
.
I=
dz
dz
dz
We differentiate Eq. (3.17) by making liberal use of Eq. (3.16) to obtain
∂F
= Re ZI + x2 Im ZI
∂x1
∂2 F
= Re ZI + x2 Im Z¢I
∂x12
48
Elements of Fracture Mechanics
∂4 F
= Re Z¢¢I + x2 Im Z¢¢¢
I
∂x14
∂F
= – Im ZI + x2 Re ZI + Im ZI = x2 Re ZI
∂x2
∂2 F
∂x22
∂3 F
∂x23
(3.18)
= – x2 Im Z¢I + Re ZI
= – x2 Re Z¢¢I – Im Z¢I – Im Z¢I = – x2 Re Z¢¢
I – 2 Im Z¢I
∂4 F
= x2 Im Z¢¢¢
I – Re Z¢¢
I – 2Re Z¢¢
I = x2 lm Z¢¢¢
I – 3 Re Z¢¢
I
∂x24
∂2 F
= x2 Re Z¢I
∂x1∂x2
∂4 F
= – x2 Im Z¢¢¢
I + Re Z¢¢
I
∂x12∂x 22
Substituting in the biharmonic Eq. (3.15a) to have the left hand side as
Re Z¢¢
I + x 2 Im Z¢¢¢
I – 2x 2 Im Z¢¢¢
I + 2Re Z¢¢
I + x2 Im Z¢¢¢
I – 3Re Z¢¢
I Æ 0
The differential equation is thus satisfied. Substitution of the relevant partial derivatives of
Eq. (3.18) in Eq. (3.14) leads to
s 11 = ReZI – x2 Im Z¢I
(3.19a)
s 22 = ReZI + x2 Im Z¢I
(3.19b)
s12 = – x2 Re Z¢I
(3.19c)
In order to solve a given problem, the proper form of the Westergaard function ZI(z) is chosen
such that the stress components, determined through Eq. (3.19), satisfy all the boundary conditions.
Once such a function is obtained, the stress field in the vicinity of the crack tip can be easily
obtained through Eq. (3.19). The Westergaard function does not solve a problem completely; it
only aids in solving a problem. We still have to guess the form of the complex function ZI(z) in a
specific problem.
To determine the displacement field (u1, u2) in the vicinity of the crack tip, we convert the
stress field to the strain field with the help of appropriate stress-strain relations (Eq. (3.10) for
plane stress problems and Eq. (3.12) for plane strain cases). Strains are then integrated for
determining displacements.
We first take the case of plane stress. Substituting Eqs (3.19a–c) in Eqs (3.10a–c), we have
∂ u1
1
= e11 = ÈÎ( Re ZI - x2 Im ZI¢ ) - n ( Re ZI + x2 Im ZI¢ ) ˘˚
∂x1
E
(3.20a)
Stress Intensity Factor
∂ u2
1
= e22 = ÈÎ( Re ZI + x2 Im ZI¢ ) - n ( Re ZI - x2 Im ZI¢ ) ˘˚
∂x 2
E
Re ZI¢
1 È ∂u1 ∂u2 ˘
+
Í
˙ = e12 = – x2 2m
2 Î ∂x2 ∂x1 ˚
49
(3.20b)
(3.20c)
Rearranging terms and making use of the relation m = E/[2(1 + n )], we simplify the first two
equations to
˘
∂ u1
1 È (1 - n )
=
Re ZI - x2 Im ZI¢ ˙
Í
∂x1
2m Î (1 + n )
˚
˘
∂ u2
1 È (1 - n )
=
Re ZI + x2 Im ZI¢ ˙
Í
∂x 2
2m Î (1 + n )
˚
We integrate both the partial differential equations using relations of Eq. (3.16) to obtain:
u1 =
˘
1 È (1 - n )
Re ZI - x2 Im ZI ˙ + f (x2)
Í
2m Î (1 + n )
˚
(3.21a)
u2 =
1
2m
È 2
˘
Im ZI - x2 Re ZI ˙ + g(x1)
Í
Î (1 + n )
˚
(3.21b)
where, f(x2) is the constant of integration which can be the function of x2 only. Similarly g(x1) is
the function of x1 only. We shall be showing that for the fracture mechanics problems, function
f (x2) and g(x1) can be equated to zero without losing any generality. To prove it, we substitute
Eqs (3.21a and b) into Eq. (3.20c) to find that all terms involving ZI cancel each other, leaving only
the following equation:
∂f (x2 )
∂g(x1 )
+
=0
∂ x2
∂x1
The equation is rearranged as
∂f (x2 )
∂g(x1 )
=–
=A
∂x2
∂x1
where A is a constant. On integration,
f (x2) = Ax2 + B
g(x2) = – Ax1 + C
Again B and C are constants of integration. If we substitute these equations into Eqs (3.21a and b)
and study them carefully, we note that all the points of the component are displaced by the same
distance, given by ul = B and u2= C. Therefore, B and C correspond to rigid body translation and
they may be set to zero without losing any generality. Furthermore, rigid body rotation w is
governed by the equation
w=
1 È ∂u1 ∂u2 ˘
Í
˙
2 Î ∂x2 ∂x1 ˚
50
Elements of Fracture Mechanics
and the contribution of function f and g towards rigid body rotation becomes
1
(A + A) = A
2
Therefore, A corresponds to a rigid body rotation of the component and may be set equal to zero
without losing any generality. Thus, for the plane stress cases,
w=
u1 =
˘
1 È (1 - n )
Re ZI - x2 Im ZI ˙
Í
2m Î (1 + n )
˚
(3.22a)
u2 =
˘
1 È 2
Im ZI - x2 Re ZI ˙
Í
2m Î (1 + n )
˚
(3.22b)
For plane strain problems, we substitute Eqs (3.19a–c) in Eq. (3.12) to obtain
∂ u1
1
È(1 - n ¢ ) Re ZI - (1 + n ¢ ) x2 Im ZI¢ ˘
=
˚
E¢ Î
∂x1
∂ u2
1
È(1 - n ¢ ) Re ZI + (1 + n ¢ ) x2 Im ZI¢ ˘
=
˚
∂x 2
E¢ Î
Substituting the values of E¢ and n ¢ [Eqs (3.11d and e)] in the above equations and then
integrating them, we obtain the results for the plane strain cases as:
u1 =
1
È(1 - 2n ) Re ZI - x2 Im ZI ˘
˚
2m Î
(3.23a)
1
È 2 (1 - n ) Im ZI - x2 Re ZI ˘
(3.23b)
˚
2m Î
Thus, we have expressed all the stress and displacement components in terms of the
Westergaard function ZI, which still is an unknown function.
The Westergaard function does not solve a problem completely. The function solves it to a
stage from where we have a much better chance to guess the form of ZI by looking at the boundary
conditions of a problem. The Westergaard approach comes out handy for the problems of an
infinite plate because we do not have to concentrate much on satisfying boundary conditions of
far field stress.
Now, we take up the problem of an infinite plate with through-the-thickness crack length of 2a,
loaded under a biaxial field of stress s as shown in Fig. 3.6. Usually, if the exterior dimensions of
the plate are much larger than the crack length, the plate is considered to be infinite. The boundary
conditions that should be met while choosing the Westergaard function are:
u2 =
(i) At the crack tip
s22 = •
(ii) On cracked surfaces (x2 = 0, – a < x1 < a)
s 22 = 0,
s12 = 0
(iii) Far away from the crack (large |z|)
s 11 = s,
s22 = s,
s12 = 0
Stress Intensity Factor
51
While guessing the form of the Westergaard function for this problem we study Eq. (3.19) in
light of the boundary conditions. The function which satisfies all the boundary conditions is
ZI(z) =
sz
1/2
(z - a)
1/2
(z + a)
=
sz
(3.24)
2
(z - a2 )1/2
s
x2
COD
s
s
x1
2a
s
Fig. 3.6
A centre-crack in an infinite plate subjected to a biaxial stress field
Now we shall check whether the form of ZI(z) satisfies the boundary conditions. For checking the
first boundary condition, using Eq. (3.19b), we find s 22 for x1 ≥ 0 and x2 = 0 as
s 22 =
(x12
s x1
- a 2 )1/2
Clearly for |x1|Æ a, the stress component s 22 tends towards infinity, thus satisfying the first
boundary condition. On cracked surfaces with x2 negligibly small ( ; 0), ZI is simplified to
ZI (z) =
s x1
(x12
(3.25)
- a2 )1/2
Then, the stress components are
s 22 = Re ZI
s 12 = 0.
ZI is imaginary on the cracked surfaces because x1 lies between – a and a, making Re ZI = 0. Thus
the second boundary condition on the cracked surfaces is satisfied. In order to check the third
boundary condition, we take |z|Æ • in Eq. (3.24) to have
ZI (z) =
leading to ZI (z) = s for
s
(1 - a
a
<< 1 .
z
2
/z
)
2 1/2
=
s
(1 - (a / z ) e )
2
2
-2iq
1/2
52
Elements of Fracture Mechanics
The differentiation of Eq. (3.24) gives
Z¢I (z) = –
s a2
(z
2
- a2
)
(3.26)
3/2
which gives Z¢I (z) = 0 when |z|Æ •. Substitution of the limiting values of ZI(z) and Z¢I (z) in
Eq. (3.19) leads to s11 = s, s22 = s and s12 = 0, thus satisfying the far field boundary conditions.
Therefore, the chosen form of ZI(z) is the correct solution to the problem.
For determining the stress field near the crack tip, it is convenient to transform the origin from
the centre of the crack to its tip. It is done by the transformation
z = a + zo
(3.27)
where zo is measured from the crack tip. ZI then becomes
ZI (z o) =
s ( a + zo )
1/2
( a + zo - a)
1/2
( a + zo + a)
=
s a [1 + zo / a]
(2zo )1/2 [1 + zo /2 a)1/2
(3.28)
Since |zo|<< a in the vicinity of the crack tip, the above equation is simplified to the approximate
relation
ZI (zo) ;
s pa
=
1/2
(2p zo )
KI
(2p zo )1/2
(3.29)
Expressing zo in polar coordinates as zo = r(cosq + i sinq ), it is modified to
ZI =
q
qˆ
Ê
ÁË cos - i sin ˜¯
2
2
KI
(2p r )
1/2
(3.30)
Similarly, the transformation z = zo + a changes Eq. (3.26) to
Z¢I (z o) = –
=
s a2
zo3/2 ( zo + 2 a)
- KI
2 ( 2p )
1/2 3/2
r
3/2
;-
s a2
zo3/2 ( 2 a)
3/2
3q
3q ˆ
Ê
- i sin ˜
ÁË cos
2
2¯
(3.31)
Substituting ZI (zo) and Z¢I (zo) in Eq. (3.19) and realizing x2 = r sin q , we obtain
s 11 =
=
KI
(2p r)
1/2
KI
(2p r )
1/2
cos
KI
q r sin q
3q
sin
1/2 3/2
2
2 ( 2p ) r
2
cos
qÊ
q
3q ˆ
ÁË 1 - sin sin ˜¯
2
2
2
(3.32a)
cos
qÊ
q
3q ˆ
ÁË 1 + sin sin ˜¯
2
2
2
(3.32b)
Similarly,
s 22 =
KI
(2p r)
1/2
Stress Intensity Factor
s 12 =
KI
sin
(2p r )
1/2
q
q
3q
cos
cos
2
2
2
53
(3.32c)
For determining the displacement field in the vicinity of the crack tip, ZI (zo) is obtained from
Eq. (3.29) as
1/ 2
1/ 2
2
ZI (zo) =  
π 
KI
 2r 
zo =  
π 


KI  cos
θ
θ
+ i sin 
2
2
Now, it is a simple matter of substituting ZI (zo) and ZI (zo) in Eq. (3.22) and simplifying it to
obtain the displacement field in the vicinity of the crack tip for the plane stress as:
KI Ê r ˆ
Á ˜
m Ë 2p ¯
1/2
u1 =
KI Ê r ˆ
Á ˜
m Ë 2p ¯
1/2
u2 =
cos
q Ê (1 - n )
qˆ
+ sin 2 ˜
Á
2 Ë (1 + n )
2¯
(3.33a)
sin
qÊ 2
qˆ
- cos2 ˜
Á
2 Ë 1 +n
2¯
(3.33b)
For the plane strain, substitution of ZI (zo) and ZI (zo) in Eqs (3.23a and b) gives:
KI
m
Ê r ˆ
ÁË ˜¯
2p
1/2
u1=
KI Ê r ˆ
Á ˜
m Ë 2p ¯
1/2
u2 =
cos
qÊ
2q ˆ
ÁË 1 - 2n + sin ˜¯
2
2
(3.34a)
sin
qÊ
2 qˆ
Á 2 - 2n - cos ˜¯
2Ë
2
(3.34b)
Unlike the stress components, the displacement components are finite and there is no square
root singularity. One should not get misled by noting that u1 and u2 keep on increasing with
increasing r; the above solution is valid only in the close vicinity of the crack tip. Also, note that u1
does not depend on the sign of q ; it is expected because Mode I problem is symmetric about the
crack plane. However, as expected, u2 changes its sign as q is replaced by – q.
The distance between the two crack faces, known as the crack opening displacement (COD), is
useful for carrying out experiments and for defining the parameter, crack tip opening displacement
(CTOD) which is introduced in Chapter 7. COD of a centre-cracked plate when subjected to plane
stress loading (Fig. 3.6) will be obtained by invoking Eq. (3.22b) for x2 = 0. For this problem ZI
and ZI simplify to
ZI =
sz
(z
2
)
2 1/2
-a
=
ZI = s (z2 – a2)1/2 = is
- is x1
a2 - x12
a2 - x12
Substituting ZI and ZI in Eq. (3.22b) and realizing the shear modulus as m = E/2(1 + n) we obtain
COD = 2u2 = 2 ¥
˘
1 È 2s
a2 - x12 ˙
Í
2m ÎÍ (1 + v)
˚˙
(3.35)
54
Elements of Fracture Mechanics
4s
a2 - x12
E
with the maximum crack opening equal to 4s a/E at x1 = 0.
In fracture mechanics problems, if Z(zo) is expressed with the crack tip as origin, zo appears
in the denominator of all the three modes. Another formal definition of K [3.7], which is an
alternative yet equivalent to the two definitions already presented through Eqs (3.3) and (3.4), is
=
2p zo Z(zo)
K=
(3.36)
zo Æ 0
If this definition is applied to Eq. (3.29), we have the familiar result, KI = s p a .
We realize that the simple expression of Eq. (3.29) was obtained by neglecting some terms from
Eq. (3.28). Some of us may, at this stage, like to split hairs and be interested in knowing how
approximate the results are for the relation of Eq. (3.29). To have some idea, we would explore
how s 22 differs from the correct solution on x2 = 0 plane. For the biaxial case under consideration,
the rigorous solution of Eq. (3.24) is simplified for x2 = 0 to
ZI (z) =
(x
s x1
- a2
)
1/2
Substituting in Eq. (3.19b) leads to
x1
s 22
=
s
x12 - a2
)
1/2
2
1
(
for |x1|> a.
The origin is moved to the crack tip by the transformation x1 = a + xo and, with some manipulation, we obtain
(1 + xo / a) Ê 1 + xo ˆ -1/2
s 22
=
Á
˜
s
(2 xo / a)1/2 Ë 2 a ¯
The approximate value of s22 is obtained from Eq. (3.32b) for q = 0 as
1
s 22
=
s
(2xo / a)1/2
The percentage difference between the two values, rigorous and approximate (labeled as ‘no
approximation’ and ‘one term’ respectively in Fig. 3.7), increases with the increase in distance
from the origin on the crack plane. At the distance where xo/a = 0.15, the difference is 9.8%.
The approximate relation of Eq. (3.29) can be improved by retaining higher terms in the
expansion of (1 + zo/2a)–1/2 of Eq. (3.28). In fact, the form of ZI (zo) changes to
ZI =
KI
(2p zo )
1/2
+ A1 z1/2
+ A2 zo3/2 + A3 zo5/2 + L
o
where A1, A2, A3, etc. are real numbers. If we retain the first two terms, s 22 is modified to
s 22 =
KI
(2p ro )
1/2
cos
q
2
q
3q 3r Ê
Ê
2 q ˆˆ
ÁË 1 + sin 2 sin 2 + 4 a ÁË 1 - sin 2 ˜¯ ˜¯
(3.37)
Stress Intensity Factor
55
3.5
s22 /s
3.0
2.5
No approximation
Two terms
2.0
One term
1.5
0.00
0.05
0.10
x0/a
0.15
0.20
Fig. 3.7 s22 on the crack plane for biaxial loading
In Fig. 3.7 the solution of 'two terms' is also plotted for s 22 from the above results for q = 0. It
shows that the inclusion of one more term considerably reduces the difference between the rigorous
and approximate values. For the biaxial loading problem, Fig. 3.8(a) shows the photo-elastic fringes
(simulated on a computer) in the approximate case of using only the first term in Eq. (3.37). The
lobes of the fringes are almost normal to the crack plane. Figure 3.8(b) shows the fringes when
two terms are used and it is evident that the lobes of the fringes are inclined. This result is close to
the fringes observed in real life cases.
(a)
(b)
Fig. 3.8 Photoelastic fringes for the biaxial loading case with analysis of (a) one term, and
(b) two terms (Courtesy: Dr K. Ramesh, Dept. of Applied Mechanics, IIT Madras)
56
Elements of Fracture Mechanics
The stress field [Eq. (3.32)] developed in this section for a biaxially loaded plate is usually
claimed to be the stress field of an uniaxially loaded plate (Secs 3.2 and 3.3). What is the
contribution of the far field stress s11 = s ? It does not cause any substantial change in the stress
field in the vicinity of the crack tip because s11 does not open up the crack. Invoking the principle
of superposition, we can separate the biaxial stress (Fig. 3.9) to two configurations, (a) and (b).
Configuration (a) is of importance to us; however, a simple solution to it is still not available.
Configuration (b) does not try to open the crack but it does modify the stress field to a certain
extent near the crack tip. However, the solution to configuration (b) is not simple and we usually
neglect its effect in engineering solutions to fracture mechanics.
s
s
s
s
s
s
=
+
a
s
s
Fig. 3.9
b
Separating biaxial loading case to configs (a) and (b)
We close this subsection by realizing that the exact Westergaard solution is not available for a
centre crack in a plate loaded uniaxially in the direction normal to the crack plane. However, a
simple Westergaard function is available for biaxially loaded plates whose stress field in the
vicinity of the crack tip is not substantially different from the case of an uniaxially loaded plate.
Therefore, for most of the practical purposes, the solution of a biaxially loaded plate is employed
for both uniaxially and biaxially loaded problems.
3.5.2 Mode II (Sliding Mode)
A centre-cracked problem in an infinite plate for Mode II loading is considered, as shown in
Fig. 3.3(a). It has been found that the following expression of the Airy Stress Function f is convenient:
F = – x2 Re ZII
(3.38)
where, ZII is a complex function and its form would be so chosen such that all the boundary
conditions are satisfied. One can easily show that the above expression for F satisfies the
biharmonic equation and the stress field is given by
s11 = 2 Im ZII + x2 Re Z ¢II
(3.39a)
s22 = – x2 Re Z ¢II
(3.39b)
s12 = Re ZII – x2 Im Z ¢II
(3.39c)
Stress Intensity Factor
57
Further, stress components are converted to strain components and then integrated to yield for
plane strain cases as
1
È 2 (1 - v ) Im Z II + x2 Re ZII ˘
u1 =
(3.40a)
˚
2m Î
u2 =
1
È -1 (1 - 2v ) Re ZII - x 2 Im ZII ˘˚
2m Î
(3.40b)
For the case of Mode II, as shown in Fig. 3.3, the Westergaard function is given as:
ZII =
tz
(z
2
- a2
(3.41)
)
1/2
which satisfies all the boundary conditions. They are similar to boundary conditions of the
Mode I problem [Fig. 3.2(a)]. Like in the case of Mode I, we transform the origin to the crack tip
by the transformation relation z = a + zo and obtain the simplified but approximate relation as
t (p a)
1/2
ZII =
(3.42)
(2p zo )1/2
Invoking the definition of the stress intensity factor [Eq. (3.36)], we obtain:
KII = t
pa
Substitution of ZII and its derivative in Eqs (3.39a–c), leads to:
s11 =
s22 =
s12 =
- K II
1/2
sin
qÊ
q
3q ˆ
ÁË 2 + cos cos ˜¯
2
2
2
1/2
cos
q
q
3q
sin
cos
2
2
2
cos
qÊ
q
3q ˆ
ÁË 1 - sin sin ˜¯
2
2
2
(2p r )
KII
(2p r )
KII
(2p r )
1/2
(3.43)
Substitution of ZII and ZII in Eqs (3.40a and b) gives the displacement field for a plane strain
problem as
1/2
qÊ
qˆ
K Ê r ˆ
u1 = II Á ˜ sin Á 2 - 2n + cos 2 ˜
(3.44a)
m Ë 2p ¯
2Ë
2¯
u2 =
KII Ê r ˆ
Á ˜
m Ë 2p ¯
1/2
qˆÊ
Ê
2qˆ
ÁË - cos ˜¯ ÁË 1 - 2n - sin ˜¯
2
2
(3.44b)
With the similar procedure, the displacement field for a plane stress field can be obtained as
KII
m
Ê r ˆ
ÁË ˜¯
2p
1/2
u1 =
KII
m
Ê r ˆ
ÁË ˜¯
2p
1/2
u2 =
qÊ 2
qˆ
+ cos2 ˜
sin Á
2Ë1+n
2¯
(3.45a)
qˆÊ1-n
qˆ
Ê
- sin 2 ˜
ÁË - cos ˜¯ Á
2 Ë1 +n
2¯
(3.45b)
58
Elements of Fracture Mechanics
3.5.3 Mode III (Tearing Mode)
A large plate with a centre crack, subjected to a far field shear stress s23 = t (Fig. 3.4), is considered.
For this case of Mode III,
u1 = 0
u2 = 0
u3 = w(x1, x2).
For this displacement field, components of strain tensor turn out to be
e11 = e22 = e33 = e12 = 0
1 ∂w
1 ∂w
; e23 =
2 ∂x1
2 ∂ x2
s11 = s22 = s33 = s12 = 0
e13 =
leading to,
s13 = 2m e13 = m
∂w
∂x1
(3.46a)
∂w
(3.46b)
∂x2
It should be kept in mind that the problem of Mode III is not a case of plane stress or plane
strain. It is considerably simpler because many components of displacement, stress and strain are
zero. We would, therefore, not be using the biharmonic equation. Instead, the problem will be
solved with displacement component w as the dependent variable and then there is no need to
worry for compatible conditions any more. Out of the three equilibrium equations, only the last
one provides the non-trivial equation,
s23 = 2m e23 = m
∂s 13
∂ó 23
+
= 0.
∂x1
∂x2
Converting it into displacement components using Eqs (3.46a and b)
∂2w
∂2w
+
=0
∂x12
∂x22
which is the well known Laplace equation and is also written as
—2 w = 0
The Westergaard approach is also applicable to this differential equation by choosing w in the
form of
1
w=
Im ZIII
(3.47)
m
where ZIII is a complex function of variable z. Again one can show that the above expression
satisfies the governing differential equation. Also, substituting w in Eqs (3.46a and b), we obtain
stress components as
s13 = Im Z ¢III
(3.48a)
s23 = Re Z ¢III
(3.48b)
Stress Intensity Factor
59
The form
Z ¢III =
ôz
(z
2
- a2
)
1/ 2
satisfies all the boundary conditions. Z¢III is transformed to the origin at the crack with relation
z = a+ zo and after neglecting the small terms, we obtain:
Z ¢III =
t a
(2zo )1/2
.
(3.49)
Using t ða = KIII and expressing zo in polar coordinates, the equation is simplified to
Z ¢III =
KIII
(2p r)
1/2
q
qˆ
Ê
ÁË cos - i sin ˜¯ .
2
2
(3.50)
Substituting Z ¢III in Eq. (3.48), we obtain the stress field as
s 13 = –
s 23 =
KIII
( 2p r)
1/2
KIII
(2p r)
1/2
sin
cos
q
2
q
2
(3.51b)
p a = KIII, we have
Integrating Eq. (3.49) and substituting t
Ê 2ˆ
ZIII = Á ˜
Ëp¯
1/2
(3.51a)
KIII ( zo )
1/2
Ê 2r ˆ
= KIII Á ˜
Ëp¯
1/2
q
qˆ
Ê
ÁË cos + i sin ˜¯
2
2
which on substitution in Eq. (3.47), gives the displacement field as
u3 =
with
KIII Ê 2r ˆ
Á ˜
m Ëp¯
u1 = 0
and
1/2
sin
q
2
u2 = 0
3.6 CONCLUDING R EMARKS
We determined the stress and displacement fields around a crack tip in infinite plates for all the
three modes. These results are quite useful to a designer, because in many practical applications
a crack in a plate is quite small in comparison to the lateral extent of the plate.
The Westergaard's approach adopted in this chapter is not the general way of solving problems.
In fact, problems of infinite plates are easier to solve because stress fields far away from the crack
are simple; we mainly focus our attention to the cracked faces and with some luck we may be able
to guess a solution, which satisfies all the boundary conditions. In the case of work-components,
60
Elements of Fracture Mechanics
which have their edge/edges close to the crack, the job of guessing a Westergaard function, which
not only satisfies conditions at cracked faces but also meets the requirements at the edges, becomes
a difficult task. Problems dealing with bodies of finite dimensions will be taken up in Chapter 4.
QUESTIONS
1. What is a singularity? What kind of singularity describes a stress field near the vicinity of
a crack tip in LEFM? Is it expected to be different for elastic-plastic fracture mechanics?
2. Stress field is the same for plane stress and plane strain problems. Why is it not so for
displacement fields?
3. In problems of plates, stress components are expressed in the Cartesian coordinate
system whereas the location at which stress is considered is defined in polar coordinates.
Why is such a mixed approach adopted?
4. For many problems of practical applications, solutions of infinite plates are applicable.
Justify the statement.
5. Displacement near the crack tip is determined by integrating strain components. Why
do we equate integration constants to zero?
6. Mode I case has been solved for a biaxial case and its stress and displacement fields are
taken to be approximately the same as of an uniaxial case. Justify.
7. Why do we not use biharmonic equation to solve Mode III problems for a centre crack in
an infinite plate?
PROBLEMS
1. Show that F = – x2 Re ZII , chosen for the Mode II problem, satisfies the biharmonic equation.
Determine stress components and displacement components (plane stress) in terms of ZII.
2. For a centre crack in an infinite plate loaded in Mode II, determine stress components and
displacement components (plane stress) near the vicinity of a crack tip in terms of KII.
1
3. Show that w =
Im ZIII , chosen for the Mode III problem for a centre-cracked infinite
m
plate, satisfies the Laplace equation — 2w = 0. Determine stress components and all
displacement components in terms of ZIII. Also, determine stress and displacement fields
in the vicinity of the crack tip in terms of KIII.
4. In a large plate, a crack of length 2a is inclined with an angle a with xl-axis (Fig. 3.10). The
plate is loaded in x2 direction with s 22 = s.
(i) Find the stress intensity factors.
(ii) For s = 80 MPa, 2a = 20 mm and a = 30°, determine KI and KII.
5. Determine stress at point H of Problem 4, if r = 1 mm and q = 45°.
6. Determine the critical crack length in a centered-cracked plate, loaded in Mode I, if critical
stress intensity factor KIc = 60 MPa m and far field stress is 120 MPa.
7. Determine stress components (srr, sq q , srq ) and displacement components (ur, uq ) in polar
coordinates for plane stress of Mode I.
Stress Intensity Factor
61
s
x2
r
q
H
a
x1
2a
s
Fig. 3.10
The figure of Problem 4
REFERENCES
3.1 Irwin, G.R. (1958). Fracture, Handbuch der Physik, S. Flugge (ed.), Springer–Verlag, Berlin,
Vol. VI, pp. 551–590.
3.2 Kanninen, M.F. and Popelar, C.H. (1985). Advanced Fracture Mechanics, Oxford University
Press, New York.
3.3 Sokolnikoff, I.S. (1956). Mathematical Theory of Elasticity, McGraw-Hill Book Company,
New York.
3.4 Fung, Y.C. (1965). Foundation of Solid Mechanics, Prentice-Hall Inc., Englewood, New Jersey.
3.5 Westergaard, H.M. (1939). “Bearing Pressures and Cracks,” Journal of Applied Mechanics,
61, pp. A49–53.
3.6 Carrier, G.F., Krook, M. and Pearson, C.E. (1966). Functions of a Complex Variable,
McGraw-Hill Book Company, New York.
3.7 Gdoutos, E.E. (2005). Fracture Mechanics—An Introduction, Springer, The Netherland.
3.8 Anderson, T.L. (2004). Fracture Mechanics: Fundamentals and Applications, CRC, Press-Book.
3.9 Sanford, R.J. (2003). Principles of Fracture Mechanics, Prentice Hall, Upper Saddle River.
3.10 Janssen, M., Zuidema, J. & Wanhill, R.J.H. (2004). Fracture Mechanics, Spon Press, Abingdon.
3.11 Ramesh, K. (2007). e-Book on Engineering Fracture Mechanics, IIT Madras,
URL: http//apm.iitm.ac.in/smlab/kramesh/book_4.htm.
Chapter
4
SIF of More Complex Cases
Models are useful, but without exception, they have blind spots built into them.
Deepak Chopra
4.1 OTHER APPLICATIONS OF WESTERGAARD APPROACH
We have so far solved the basic problems of Modes I, II and III in the previous chapter through
the approach made available by Westergaard. The chapter defined stress intensity factor and
determined stress and displacement fields near a tip of a crack in an infinite plate loaded with a
far field stress. In this section, we would take up some more problems which are also important
for the practical cases.
4.1.1 Wedge Loads on Cracked Surfaces
Consider the problem of two wedge loads P (per unit length), acting symmetrically on each cracked
x2
a
a
P
P
x1
P
s
P
s
Fig. 4.1 Symmetrically applied wedge loads on the surfaces of a crack
SIF of More Complex Cases
63
surface (Fig. 4.1) at a distance s from the centre of the crack. The problem has applications to riveted
joints. For this case, the Westergaard function is known to be
ZI =
(
2Pz a2 - s 2
(
p z2 - s2
) (z
2 P (p a)
KI =
p a2 - s 2
2
)
1/2
- a2
)
1/2
1/2
which leads to
(
(4.1)
)
1/2
Another case of wedge loads on one side only (Fig. 4.2) is more general, and stress intensity
factors are known to be [4.1]:
K AI
K BI
=
P Èa + s˘
Í
˙
p a Îa - s˚
=
P
pa
ÈaÍ
Îa+
s˘
˙
s˚
1/2
(4.2a)
1/2
.
(4.2b)
x2
a
a
P
B
A
x1
P
s
Fig. 4.2 A pair of wedge loads at distance s on the surfaces of a crack
These results are also found to be very useful in solving problems of distributed pressure on
the crack surfaces by constructing Green's functions. To explore the approach of Green's function,
we consider the problem of a triangular pressure distribution at the cracked surfaces as shown in
Fig. 4.3 with maximum pressure p0 acting at the centre. This problem is solved by invoking the
solution to the symmetric line loads acting on the faces of the crack [Eq. (4.1)]. Consider two thin
slices, each of width ds and at a distance s from the origin, as shown in the figure. Then, the load
on one slice is given by
dP =
( a - s)
a
p0ds
64
Elements of Fracture Mechanics
x2
ds
s
ds
s
p0
x1
p0
a
a
Fig. 4.3 Triangular pressure distribution acting at the surfaces of a crack
which contributes towards stress intensity factor as
dKI =
2 ( a - s) p0 p a ds
(
ap a2 - s 2
)
1/2
.
Now, to determine the effect of pressure, the equation is integrated to have the form
( a - s) ds
2
p0 p a Ú
2
2 1/2
p
0 a a -s
a
KI =
(
)
which on integration gives
KI = (1 – 2/p )p0 p a .
Note that we have considered a simple example for demonstrating the approach of Green's
function; but problems having any variation of p(x) on the crack surfaces can be solved through
Eq. (4.1) for symmetric pressure, or through Eq. (4.2) for asymmetric pressure. If p(x) is a simple
distribution, one may be able to integrate to obtain a closed form solution; otherwise, one can
always resort to numerical techniques to solve the problem.
4.1.2 Collinear Cracks in an Infinitely Long Strip
A classic problem in fracture mechanics is of collinear cracks in an infinitely long strip as shown
in Fig. 4.4. Identical cracks, each of length 2a, are separated by a distance W. The geometry of this
problem is not usually encountered in practical cases and, therefore, one may think that the
problem is solved just for the sake of academia. This is not the case, because the problem acts as
a stepping stone to several real life problems dealing with finite size plates. In this section, we
shall find the solution of this problem, but in subsequent sections of this chapter, appropriate
portions will be cut out from this strip to solve problems encountered in several engineering
applications of importance. The Westergaard function ZI for the problem [4.2] is known to be
SIF of More Complex Cases
65
X2
s
A
C
W
B
W
X1
2a
2a
A¢
C¢
2a
B¢
s
Fig. 4.4 Collinear cracks in an infinitely long strip
s sin (p z /W )
ZI =
(4.3)
È sin (p z /W ) - sin 2 (p a /W ) ˘ 1/2
Î
˚
where, the origin is at the center of a crack. Transforming axes to the tip of the crack by relation
z = a + zo leads to
2
sin
pz
pa
pa
Êp z ˆ
Êp z ˆ
= sin Á o ˜ cos
+ cos Á o ˜ sin
Ë W ¯
Ë W ¯
W
W
W
(4.4)
Since we have confined the analysis to the close proximity of the crack tip (|z|<< a),
p zo
Êpz ˆ
sin Á o ˜ ª
Ë W ¯
W
p zo
ª1
W
Simplifying Eq. (4.4) with the above approximations and substituting in Eq. (4.3), we obtain
cos
ZI =
pa
p a˘
Èp z
s Í o cos
+ sin
W
W ˙˚
Î W
È p 2 zo2
pa
p a˘
2p zo
2 pa
+
cos
sin
Í 2 cos
˙
W
W
W
W˚
ÎW
1/2
Since zo/W is a small number, the first term of the numerator as well as of the denominator are
negligible, leading to
66
Elements of Fracture Mechanics
ZI =
p aˆ
Ê
s Á sin
˜
Ë
W¯
1/2
p aˆ
Ê 2p zo
ÁË W cos W ˜¯
1/2
Using basic definition of KI [Eq. (3.36)], we obtain
KI = (2p zo)1/2ZI
zo Æ 0
=
p aˆ
Ê
s Á sin
Ë
W ˜¯
1/2
p aˆ
Ê 1
ÁË cos
˜
W
W¯
1/2
Manipulating the equation, so as to have the familiar s p a in the numerator, we have
p aˆ
Ê
tan
Á
W ˜
KI = s p a Á
pa ˜
Á
˜
Ë W ¯
1/2
.
(4.5)
It is worth noting that the expression of Eq. (4.5) is only an approximate relation. For W > > a, we
obtain KI = s
p a , which is the same as that of an infinite plate with one crack.
4.2 APPLICATION
OF THE
P RINCIPLE
OF
SUPERPOSITION
The principle of superposition can be applied to identical geometries for linear elastic bodies.
Since only elastic bodies are considered in this chapter, the principle can be exploited to solve
problems of fracture mechanics.
Take an example of a centre-cracked plate with two different loads, s as a far field stress and
line loads P at the cracked faces as shown in the Config. (s) of Fig. 4.5. Invoking the principle of
superposition, stress at any point H in the vicinity of the crack tip is given by the sum of the stress
of Config. (m) and Config. (n); that is,
m
n
s22 = s 22
+ s 22
m
n
Substituting the values of s 22
and s 22
from Eq. (3.32b), we get
s 22 =
=
KIm
KIn
qÊ
q
qÊ
q
3q ˆ
3q ˆ
cos Á 1 + sin sin ˜ +
cos Á 1 + sin sin ˜
2Ë
2
2¯
2Ë
2
2¯
2p r
2p r
(K
m
I
+ KIn
2p r
) cos q Ê 1 + sin q sin 3q ˆ
Á
2Ë
2
˜
2¯
(4.6)
SIF of More Complex Cases
s
s
P
H
.
... .
H
....
P
..
...
H
.
.
.
...
.
P
P
m
s
s
67
n
s
Fig. 4.5 Separating a complex problem into two simpler problems to exploit
the principle of superposition
which means,
KI = K mI + K nI
(4.7)
We thus conclude that when several external loads develop stress intensity factors of Mode I
on a crack, the resulting stress intensity factor is the sum of individual stress intensity factor.
However, it is clear from Eq. (4.6) that in the case of mixed modes, the stress intensity factors
cannot be added to obtain the net stress intensity factor.
4.2.1 Internal Pressure on Cracked Faces
Consider the case of a large plate with an uniform pressure p applied on the crack faces (Fig. 4.6a).
The pressure opens the crack faces and loads the specimen in Mode I. We would be determining
the resulting stress intensity factor at the crack tips. The solution to this problem can be easily
obtained through the Green's function approach using Eq. (4.1), as
KI = p p a
The same problem can also be solved using the principle of superposition. This method is
different from the approach of Green's function. In fact, several interesting and powerful tricks
are developed which can be used for solving more complex problems of fracture mechanics
through the use of the principle of superposition.
Consider an infinite plate which has no crack but is loaded by a far field stress s, shown through
the starting Config. (s) in Fig. 4.6(b). Config. (s) is equivalent to Config. (f) with a centre-crack and
the applied traction s on the cracked faces, whose magnitude must be the same as that of the far
field stress. This is because the neighboring atoms on the opposite side of the crack are no longer
held together with the help of the interatomic bonding. In place of these bonds, external traction
is applied on the cracked faces so that they do not open up under the far field stress and the stress
in the entire plate remains uniform. Now, the Config. (f) has two external loads, the far field stress
and the applied traction at the cracked faces. They are separated, keeping the geometry of the
crack and the plate same, into Config. (g) and Config. (h). Thus, the superposition of Config. (g)
and Config. (h) makes Config. (f), whose stress intensity factor is zero. Invoking the principle of
superposition, we obtain:
68
Elements of Fracture Mechanics
p
p
2a
(a)
2a
s
f
g
h
(b)
Fig. 4.6 (a) Internal pressure on the surfaces of a crack, and (b) determination of KI
by invoking the principle of superposition
g
KI + KIh = 0
g
KIh = – KI = – s p a
If the direction of the externally applied traction in Config. (h) is reversed (p = – s ), we have the
stress intensity factor of the centre-crack with internal pressure p, as
yielding
KI = p p a
4.2.2 Wedge Load at the Surface of a Crack Face
Consider the case of a line load P at one surface of the crack (Config. (s) of Fig. 4.7) and a far field
stress s on the side of the other crack face. Note that for maintaining equilibrium, P = s W. Such
a problem has practical applications in riveted joints. The Config. (s) can be separated into
Config. (h) and Config. (n). As shown in the figure, Config. (h) is further separated into Config. (l)
and Config. (m). The principle of superposition gives
KIs = KIh - KIn
= KIl + KIm - KIn
(4.8)
SIF of More Complex Cases
s
69
s
W
P
P
2a
P
P
h
s
n
s
s
s
s
P
P
l
m
P
n
s
Fig. 4.7 A wedge load on one surface of a crack and a far field stress
on the side of the other crack face
It is to be noted that loads on Config. (n) are the same as those on Config. (s), except with the
difference in the direction of the loads. When the Config. (n) in Fig. 4.7 is taken to the left hand
side of Eq. (4.8), it gets added to Config. (s), giving
2K sI = K lI + K mI
Using Eq. (4.2a) and substituting P = s W, we have
KM
I =
sW
P
=
pa
pa
The expression for K Il is well known as s p a for the infinite plate. Thus, we have
sW
K Is= s p a +
2
pa
2
4.3 CRACK
IN A
PLATE OF FINITE D IMENSIONS
In practical applications, an edge (boundary) of a component may be close to the crack tip. Since
the edge is traction free, it disturbs the stress field around the crack tip. The edge then may have
a considerable influence on the stress field in the vicinity of the crack tip and on the stress intensity
70
Elements of Fracture Mechanics
factor. The situation may require an accurate determination of the SIF. When is an edge regarded
close to the crack? If the distance of the edge from the crack tip is less than the crack length, or of
the order of the crack length, the component is of finite dimensions.
Experimentalists, who determine the toughness properties of materials, also require accurate
analysis of the SIF for a test specimen. A crack in a test specimen is fairly long (usually longer
than 10 mm) for practical considerations. The lateral dimensions of the specimen cannot be made
very large for reasons such as to save money on the material of the specimen, to reduce machining
charges, to use a test-machine of low load capacity and to minimize material handling problems.
Since free edges of a specimen have considerable influence on the stress field and the SIF, it
becomes necessary to analyze the problem accurately. Usually, SIF is expressed in the form
K = s p a f ( a /W )
where, W is the width of plate (dimension of the component from one edge to opposite edge
along the crack length) and the function f depends on a/W. For most cases, f(a/W) is written as
a series of ratio a/W.
The Westergaard approach is applicable only to a limited cases of finite dimensions. More
problems are solved by the general approach of Muskhelishvili, in which Airy's Stress Function
is expressed in terms of two complex variables. An introduction to this method is discussed in
Appendix 4A. It is worth mentioning here that the Westergaard approach is a special case of
the general Muskhelishvili approach [4.3]. If a close form solution to a problem is not available,
powerful numerical methods have been developed for determining stress intensity factor of a
crack in a given component.
We now consider the case of a common problem—a centre-cracked plate of finite dimensions
subjected to a tensile stress s which acts normal to the crack plane. The SIF is estimated using
the results of collinear cracks in an infinite strip (Fig. 4.4), by cutting a portion out at AA’ and
BB’. The separated portion is shown in Fig. 4.8, with traction on the cut faces. Since AA’ and BB’
are the planes of symmetry, shear stress on them is zero. The stress component s11 has some
distribution, shown qualitatively in the figure. In fact, the problem is somewhat similar to the
biaxial loading (Fig. 3.6), where it is argued that s11 does not change the SIF significantly. On the
same lines, the effect of s11 on cut faces may be ignored. Then, the SIF of a plate with a centrecrack is approximated to be same as that of the case of collinear cracks in an infinite strip, i.e.,
1/2
p a˘
È
Í tan W ˙
KI = s p a Í
˙
Í pa ˙
Î W ˚
However, the Mode I problem of a finite plate with a centre crack is important and has been
solved through advanced mathematical techniques and sophisticated numerical methods. It has
been found that the exact solution is more close to the form
È
Ê p aˆ ˘
KI = s p a Í sec Á
Ë W ˜¯ ˙˚
Î
1/2
SIF of More Complex Cases
71
s
A
B
x2
s11
s11
x1
2a
W
A¢
B¢
s
Fig. 4.8 Centre-cracked plate of finite dimensions
Note that for a/W < < 1, KI approaches s p a which, as expected, is the solution of an infinite
plate.
4.4 EDGE C RACKS
Edge cracks are more dangerous than interior cracks. We have already discussed in Sec. 4.3 that a
free edge close to the crack influences the stress field near the crack tip. In the case of an edge
crack, the free edge is not only close to the crack, but it intersects the crack (touches the surfaces
of the crack). Edge cracks are very commonly encountered in day-to-day life. Since they are more
dangerous, special attention is required to deal with them.
Consider an edge crack in a semi-infinite plate which is loaded by a far field stress s, as shown
in Fig. 4.9. The stress intensity factor for this case is known [4.2] to be equal to 1.12 s p a . When
the edge crack is compared with one half of the overall length of an interior crack, the value of the
SIF is about 12% more. We can justify the larger value by looking at the problem closely. The ends
of cracked faces at the free edge tend to open up more easily. This is similar to the case of a
cantilever beam whose deflection is more than the deflection of a beam supported at its two ends.
SIF for a finite plate with an edge crack can be determined through the expression given in
Appendix 4B.
The analysis for the case of edge crack is not straightforward, because the mouth of the crack
lies on the free edge and the stress field is influenced considerably by the free edge. The problem
can be solved by separating a portion CBB¢C¢ from the strip of collinear cracks (Fig. 4.4) and
invoking the principle of superposition to make the traction zero on section . However, the solution
is quite complex and is beyond the scope of this book.
72
Elements of Fracture Mechanics
s
a
s
Fig. 4.9 Edge-crack in a large plate
4.5 EMBEDDED CRACKS
So far in this chapter, we have considered through-the-thickness cracks, i.e., the crack front runs
from one face of the plate to another. But there are many cracks of practical importance where a
crack initiates at one face of the plate, but does not go all the way to the other face. The front of
such a crack is usually curved and embedded within the thickness of the plate. Embedded cracks
are usually modeled as a semi-ellipse in the literature of fracture mechanics. Figure 4.10(a) shows
an embedded crack, also known as surface crack. If the crack happens to be at the corner of the
plate (Fig. 4.10(b)), it is usually modeled as a quarter-elliptical-crack.
When a semi-elliptical crack becomes critical, it may grow along the minor axis a, or major axis
c, or along both. Later in this section, we would show that for c > a, the crack in most cases tends
to grow more into the depth and less along the lateral direction.
Surface cracks are observed in boilers, compressor-chambers, pipelines, etc., which carry high
pressure fluids or gases. Also, many structural components with embedded cracks are subjected to
flexural loads. The SIF of some important cases may be obtained from handbooks [4.2, 4.4] or other
reference books [4.1, 4.5, 4.6]. The SIF may also be determined using a FEM computer package.
It is worth noting that embedded cracks, such as those shown in Fig. 4.10a are exposed to a free
surface of the plate. The free surface influences the stress field and the SIF considerably. This is
similar to the case of the edge crack of through-the-thickness in which the free surface affects the
stress field and makes the analysis difficult (Sec. 4.4). Therefore, the analysis of an embedded
crack, with its crack face meeting the surface of the plate, is complex and one generally determines
SIF through numerical techniques. However, one way to estimate the SIF of surface cracks is by
considering the solution to an elliptical crack that is fully embedded in a plate, as shown in
Fig. 4.11. Once the solution to this problem is obtained, the SIF of the semi-elliptical crack is
approximated to be 12% higher, similar to the practice adopted for through-the-thickness edge
cracks.
SIF of More Complex Cases
73
x2
x3
s
x1
s
a
2c
a
c
s
s
(a)
(b)
Fig. 4.10 (a) A semi-elliptical surface crack, and (b) a quarter-elliptical corner crack
s
2c
2a
q
2a
2c
s
Fig. 4.11 A fully embedded elliptical crack
4.5.1 Elliptical Crack
The problem of the fully embedded elliptical crack was first presented in complete form by Irwin
in 1962 for a large plate of infinite thickness [4.7]. The SIF at a point of the elliptical crack front is
given by (Fig. 4.11)
Elements of Fracture Mechanics
74
s (p a)
1/2
KI =
I2
2
È 2
˘
Ê aˆ
2
Ísin q + Á ˜ cos q ˙
Ë c¯
ÎÍ
˚˙
1/4
(4.9)
where I2 is the elliptical integral of a second kind that depends on a/c (ratio of minor to major
axis) as defined in Fig. 4.11. I2 is given by
p /2
I2 =
Ú
0
Ê
ˆ
c 2 - a2
1
sin 2 a ˜
Á
2
c
Ë
¯
1/2
da
(4.10)
Since the evaluation of I2 is not handy for a designer, its value is generally provided through a
table or an approximate algebraic relation. We have chosen to present it through Table 4.1.
TABLE 4.1 Value of elliptical Integral I2
a/c
I2
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.000
1.016
1.1051
1.097
1.151
1.211
1.277
1.345
1.418
1.493
p/2
4.5.2 Semi-elliptical Cracks
As mentioned before, a surface crack is modeled as a half ellipse with its minor axis into the
thickness direction. The SIF of the surface crack is then 12% higher over the corresponding SIF of
the elliptical crack. Thus, the SIF at a point of the crack front of a semi-elliptical crack is
1.12s (p a)
1/2
KI =
I2
1/4
2
È 2
˘
Ê aˆ
2
q
sin
+
Í
ÁË ˜¯ cos q ˙
c
ÎÍ
˚˙
(4.11)
At the extreme end of the minor axis (q = 90°), the SIF is
K 90
I =
1.12s (p a)1/2
I2
and at the extreme end of the major axis of the ellipse (q = 0°), the SIF is given by
K 0I =
1.12s (p a)1/2
I2
Ê aˆ
ÁË ˜¯
c
1/2
It is worth noting that the segment of the crack tip, which is deep inside the material, possesses
a higher SIF. Thus, a crack tends to grow deeper into the thickness, rather than sideways on the
surface, as shown in Fig. 4.12.
Also, for a very shallow crack (a << c), I2 from Table 4.1 is close to unity and the SIF at q = 90°,
becomes
1/2
K 90
I = 1.12s (p a)
which is the same as the result of through-the-thickness edge crack of length a. In this case, the
dimension of the major axis is no longer relevant. Therefore, a shallow crack is equivalent to a
through-the-thickness edge crack of length a.
SIF of More Complex Cases
75
q
a
Crack
growth
2c
Fig. 4.12
The growth of a semi-elliptical surface-crack
The problem of the surface crack in a plate, where the crack depth a is comparable to thickness
of the plate, has also been solved through finite element methods [4.4]. The SIF at the deepest
segment of the crack front is expressed as
KI
1/2
p a)
(
= Ms
I2
(4.12)
where the factor M is 1.12 for a very small a/t, t being the plate thickness. M increases with the
increasing a/t, but it depends on two ratios, a/t and a/c. For further details, readers are referred
to a handbook on stress intensity factors [4.4].
4.5.3 Quarter or Corner Cracks
A corner crack under tensile load [Fig. 4.10(b)] is exposed to two free surfaces, each one applying
an additional correction factor on the SIF. The overall correction factor of such cracks in thick
plates has found to be about 20% higher [4.1] i.e., the largest SIF on the crack front is
KI = 1.2 p a
4.6 THE R ELATION BETWEEN GI AND KI
Energy release rate GI is a global parameter and deals with energy. On the other hand, stress
intensity factor KI is a local parameter which deals with displacement and stress fields in the
vicinity of the crack. Although the approaches are entirely different, the goal is same, i.e., to
characterize a crack. It is like measuring edible oil through either mass or volume; both methods
are employed in the Indian market. There is a relationship between the two methods. In fact, if
the oil is sold by the liter, the relation should be prescribed on the container as per the law of the
Government of India. Similarly, we should have a relation between G and K.
76
Elements of Fracture Mechanics
The relation was obtained by Irwin [4.8]. He however employed a method which is somewhat
indirect and some beginners may find the proof to be a bit difficult to accept at the first reading.
Since the two approaches are quite different, the global approach of G is made to become local
through Irwin's clever manipulation.
Consider a crack of length a which is extended by an incremental length Da (Fig. 4.13). Note
that we have the freedom to choose Da as small as we want—a point that can be exploited later. In
order to maintain clarity the extended crack and its associated parameters are referred by the
prime system; i.e., crack length a + Da is designated as a¢ and the SIF based on a¢ is denoted by K¢.
At a distance s, which is at a distance (D a – s) from the extended crack tip, the displacement of a
crack face (u2) is determined for plane stress cases from Eq. (3.33b) for q = 180° as
u2(s) =
KI¢ Ê Da - s ˆ
m ÁË 2p ˜¯
1/2
2
(1 + n )
Da
s
s
a
a
Fig. 4.13
ds
s
Closure of the crack to find relation between GI and KI
Now, each crack face in the portion Da is moved in through the distance u 2(s) with the help
of the traction s22 so that the crack faces touch each other. The magnitude of stress s 22 is
evaluated from the stress field of the unextended crack of length a, and therefore at a distance s,
it is given by
s22(s) =
KI
(2p s)1/2
Thus, the crack is closed by length Da. Irwin argued that the total elastic work required by s22 in
closing the crack is equal to the energy released. Balancing the two energies, we have
Da
GI BDa = 2
Ú
0
Bs 22 u2
ds
2
(4.13)
where B is the thickness of the plate; the integral is multiplied by 2 to account for strain energy in
both cracked faces and divided by 2 to account for the linear relationship between s22 and u2.
Substituting s22 and u2 in the equation and taking limit Da Æ 0, we have
SIF of More Complex Cases
2
GI = lim
DaÆ0 (1 + n ) mDa
Da
Ú
0
KI
KI¢ ( Da - s)
(2p s)1/2
77
1/2
(2p )1/2
ds
(4.14)
We can express
K I¢ = KI + DKI
Since D a can be chosen as small as we like, DKI can be made small enough to be neglected in
comparison to KI. Then, Eq. (4.14) is simplified to
KI2
DaÆ 0 (1 + n ) p mDa
GI = lim
Da
Ê Da - s ˆ
Ú ÁË s ˜¯
1/2
ds
0
To solve the equation, we substitute s = D a sin2a to obtain
GI =
KI2
(1 + n ) p m
p /2
Ú
2cos2 a d a
0
Solving the integral and realizing the shear modulus m = E/2(1 + n ), we obtain
KI2
(4.15a)
E
The relation is simple, but is rigorous only for brittle materials in which the components remain
elastic.
During the historical development of Fracture Mechanics, the relation of Eq. (4.15a) was
made simple at the cost of making the expression of the SIF more complex than necessary.
There is no need to have p under the square root in the expression KI = s p a . It should have
GI =
been KI = s a with GI = pK21/E. However, with so much work already carried out with p in
the definition of KI, it is extremely difficult to correct the historical mistakes. There are many
similar compromises in the history of science and technology. For example, one hour has 60
minutes, or the angle around a point is 360° in place of a convenient decimal system just because
a long time ago in Babylonia, there were sixty Gods leading to a sexagesimal number system
(based on the number sixty).
In the case of a plane strain, the relationship becomes:
KI2
E
Similar relations for Mode II and Mode III can be obtained as:
GI = (1 – n 2)
GII =
KII2
E
GII = (1 – n 2)
KII2
E
for plane stress
(4.15c)
for plane strain
(4.15d)
2
KIII
K2
= III
E
2m
In case all three modes are present, the energy of each mode is added up to GTOT, as
GTOT = GI + GII + GIII
GIII = (1 + n )
(4.15b)
(4.15e)
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Elements of Fracture Mechanics
4.7 CRITICAL STRESS INTENSITY FACTOR
Two Questions:
Why should we evaluate the SIF for a crack in a component?
How does it help a designer?
Recall that our prime goal of studying fracture mechanics is to predict whether a crack is likely to
grow or not. If the SIF of a crack approaches or exceeds an upper limit of the stress intensity factor,
the crack may grow. The upper limit is known as the critical stress intensity factor which is a material
property and is usually denoted by the symbol KIc for Mode I cases (KIIc and KIIIc for Mode II and
Mode III respectively). In order to provide a feel of stress intensity factor and the critical stress
intensity factor, an analogy is made with stress and yield stress of a solid. Stress is a parameter
which represents internal loading within the solid and yield stress is the limit on stress, beyond
which the material is regarded to have failed by many designers. Similarly, stress intensity factor is
a parameter to measure the severity of stress at the crack tip. But, critical stress intensity factor is the
limit on the SIF, such that if the SIF exceeds the critical stress intensity factor, the crack may grow.
Thus, in order to predict the growth of a crack in a component, the designer should find two
values: (i) the SIF determined through analysis for the geometry of the component, crack
configuration and applied loads and (ii) the critical SIF determined through experiments for the
material of the component. If the stress intensity factor exceeds the critical stress intensity factor,
the designer should do something, such as reducing the loads on the component, modifying the
geometry of the component, or choosing a material of higher toughness.
The SIF of some commonly encountered cases is listed in Appendix 4B. In addition, several
handbooks [2.2, 2.4] are available and the designer may find the SIF of an application under
consideration. Otherwise, he may have to determine the SIF for the problem usually through a
numerical technique.
One difficulty, faced during the experimental determination of the critical SIF for the material of
the component, is that critical SIF is found to be dependant on the thickness of a plate. In fact, the
critical SIF is independent of the thickness only in the case of a thick plate, because the plate is then
loaded in the plane strain. One question remains—what is the criterion of assuring that the plane
strain conditions prevail? The size of the plastic zone in the vicinity of the crack tip decides it. If the
plate thickness is significantly greater than the size of the plastic zone, then the conditions of plane
strain exist. Determination of the plastic zone size is discussed in detail later in Chapter 5.
Thus, we find that the critical SIF becomes a property of the material only for plane strain
cases. Therefore, in handbooks and literature, the values of critical SIFs of commonly used
materials (steel, aluminum, titanium, brass, etc.) are given for plane strain conditions.
Critical stress intensity factor for thin plates depends on the plate thickness and its value is
rarely provided as a function of thickness in literature. However, the critical SIF of a plane stress
case is higher than the corresponding value in a plane strain. A designer may find that a component
is subjected to plane stress, but critical SIF is available only for plane strain. He may safely use the
critical SIF of the plane strain because it would provide a conservative design.
In certain design problems (e.g., components of airplanes, rockets and spaceships), using the
critical SIF of a plane strain as material property may be too conservative, because structural
plates are mostly used in plane stress. Adopting a too conservative approach is against the
philosophy of engineering profession. Engineers should always strive to obtain numbers close to
SIF of More Complex Cases
79
reality. In aerospace applications, where the factor of safety is of the order of 1.1 for many
components, using material toughness properties of plane strain is likely to make the machine
heavy with poor payload. In such a situation, stress intensity factor is determined by preparing a
test-specimen of same thickness as of plates used in the actual application.
However, for most down-to-earth problems (such as the designing components of automobiles,
roof trusses, locomotive carriages, pipe lines, etc.) the conservative approach of using the critical
SIF of the plane strain may be quite practical and useful.
Companies, specially dealing with aerospace parts, do generate data for in-house use.
Unfortunately, the data is usually not placed in any open literature (handbooks, journals), so that
competitors are not able to take any advantage. Looking through a bird's eye view, many of us
think that there is unnecessary wastage of resources if the properties of a material are
experimentally determined at several places in the world. An open policy is desirable where the
companies share data freely. This will reduce the cost of testing and consequently the cost of their
products.
Critical SIF of a material depends on many factors, such as
∑ Heat treatment which controls the yield stress of the material.
∑ Speed of the crack.
∑ Temperature of the specimen.
∑ Process of manufacturing (e.g., vacuum furnaced or air melted, as cast or rolled).
∑ Orientation of the crack with respect to the grains at the crack tip.
∑ Test method.
To provide a feel to the readers about the toughness of engineering materials, representative
KIc is listed in Table 4.2 for quasi-static load conditions, at room temperature. In the table, the
process of heat treatment of alloys has been accounted for by listing the yield stress of the material
along with the KIc. For alloy steel or maraging steel, one can note from the table that KIc decreases
substantially when the yield stress is increased by heat treatment. However, KIc of a material in
Table 4.2 is only a representative value.
TABLE 4.2 Representative KIc of various materials
Material
Mild Steel
Yield stress, MPa
240
Medium Carbon Steel
Alloy Steel*
260
860
1070
1515
1850
Rotor Steel
Nuclear Reactor Steel
Maraging Steel
626
350
1770
2000
2240
KIc , MPa m
very high
( ; 220)
54
99
77
60
47
50
190
93
47
38
(Contd.)
80
Elements of Fracture Mechanics
Material
Yield stress, MPa
Stainless Steel
Aluminum
2014-T4
2014-T651
7075-T651
7178-T651
Titanium (Ti-6Al-4V)
KIc , MPa m
80–150
460
455
495
570
910
Perspex (PMMA)
PVC
Nylon
29
24
24
23
55
1.6
3.5
3.0
* 40Ni2Cr1Mo28 (IS)/EN 24 (UK)/4340 USA
4.8 BENDING AND T WISTING
OF
CRACKED P LATES
In this chapter, so far, we have discussed cases where plates are subjected only to simple tensile
or shear stress components. In such cases, stress components do not vary across the thickness of
the plate. But in several engineering applications, a plate is subjected to bending and torsional
moments and shear forces. Stress developed by a bending moment varies across the thickness,
tension on one surface and compression on another. Similarly, twisting moments and shear forces
develop stresses which vary across the thickness. Thus, the analysis is more complex.
4.8.1 Terminology of the Plate Theory
Consider a square element which has been sectioned out from a plate (Fig. 4.14). In the case of a
thin plate, s33 is negligible at all points of the plate. All the other five stress components are
developed by three moments per unit length of the plate (M1, M2 and M12) and two shear forces
per unit length (V1 and V2). M1 and M2 are bending moments, whereas M12 is a twisting moment.
If h is the thickness of the plate, all moments and shear forces can be expressed in terms of stress
components, as follows:
h /2
M1 (x1, x2) =
Ú
s 11 x3 dx3
- h/2
h /2
M2 (x1, x2) =
Ú
s 22 x3 dx3
- h/2
h /2
M12 (x1, x2) =
Ú
s 12 x3 dx3
- h/2
h /2
V1 (x1, x2) =
Ú
- h/2
s 13 dx3
(4.16)
SIF of More Complex Cases
81
x3
M21
M2
M12
x2
h
2a
V2
M21
M12
V1
M2
M1
x1
Fig. 4.14 A centre-cracked plate loaded with bending moments M1 and M2,
twisting moment M12, and shear forces V1 and V2
h /2
V2 (x1, x2) =
Ú
s 23 dx3
- h/2
For linear elastic materials in plane stress, components, s 11, s 22, s 12 vary linearly in x3 directions
and shear components s 13 and s 23 have parabolic distribution. In case moments and shear forces
are known, the stress components can be determined by the following relations:
12x3 M1
h3
12x3 M2
s 22 =
h3
12x3 M12
s 12 =
h3
s 11 =
s 13 =
2
3 È Ê 2 x3 ˆ ˘
1
Í Á
˜ ˙ V1
2h ÎÍ Ë h ¯ ˚˙
s 23 =
3 È Ê 2x3 ˆ
Í1 - Á
˜
2h ÍÎ Ë h ¯
2˘
˙ V2
˙˚
4.8.2 Through-the-Thickness Crack in a Plate
For the crack shown in Fig. 4.14, surfaces of the crack are traction free (s22 = s12 = s23 = 0). Invoking
Eq. (4.16), we obtain the following boundary conditions on the crack surfaces:
82
Elements of Fracture Mechanics
M2(x1, 0) = 0 ; M12(x1, 0) = 0 ; V2(x1, 0) = 0
The solution to this problem is complex and only approximate or numerical results are available.
It has been found convenient to express all the three moments and the two shear forces in the
vicinity of the crack tip in terms of the three moment intensity factors K1, K2 and K3. Note that
t
these SIFs are different from KI, KII and KIII earlier used in this chapter. M t1, M t2 , M12
, V 1t and V 2t ,
are evaluated in the vicinity of a crack tip (Fig. 4.14) by the expressions [4.9]
M 1t =
M 2t =
t
M 12
=
V 1t =
V 2t =
K1
(2r )
1/2
K1
(2r )
1/2
K1
(2r )
1/2
- K3
(2r )
1/2
K2
qÊ
q
qÊ
q
3q ˆ
3q ˆ
cos Á 1 - sin sin ˜ sin Á 2 + cos cos ˜
2Ë
2
2 ¯ ( 2r )1/2
2Ë
2
2¯
K2
qÊ
q
q
q
3q ˆ
3q
cos Á 1 + sin sin ˜ +
sin cos cos
1/2
2Ë
2
2 ¯ ( 2r )
2
2
2
sin
q
q
qÊ
q
K2
3q
3q ˆ
cos cos
cos Á 1 - sin sin ˜
+
2
2
2 ( 2r )1/2
2Ë
2
2¯
sin
q
2
K3
1/ 2
cos
( 2r )
θ
2
Formal definitions of K1, K2 and K3 can be stated as:
t
K1 = lim 2 r M 2 ( r , q = 0 )
rÆ0
t
K2 = lim 2 rM12 ( r , q = 0)
rÆ0
t
K3 = lim 2 rV2 ( r , q = 0 )
rÆ0
4.8.3 Bending Moment on a Centre-Cracked Plate
•
Consider a large plate with far field moments, M •1 , M •2 , and M 12
, acting on it with a crack aligned
•
with x1 axis. The effect of M 1 on the stress field near the crack tip is small. This is analogous to the
•
small effect of s 11
on the plate considered under biaxial tension (Fig. 3.6). For design purposes,
one can evaluate the SIFs from the following:
K1 = FM •2 a
K2 = yM •12 a
K3 = -
10 W
M •12 a
(1 + v) h
The large and thin plates F, Y, W, determined numerically, are shown in Figs 4.15–4.17.
SIF of More Complex Cases
1.0
0.8
F
v = 0.5
0.6
0.3
0.0
0.4
0.2
0
0
0.5
1.0
1.5
2.0
2.5
3.0
10 (h/a)
Fig. 4.15 F for a centre-crack in a thin plate [4.9]
1.0
0.8
v = 0.5
0.6
Y
0.3
0.0
0.4
0.2
0
0
0.5
1.0
1.5
2.0
2.5
3.0
10 (h/a)
Fig. 4.16 Y for a centre-crack in a thin plate [4.9]
83
84
Elements of Fracture Mechanics
0.2
v = 0.5
0.25
0.0
W 0.1
0
0
0.25
0.5
0.75
1.0
1.25
1.5
10 ( /a)
Fig. 4.17 W for a centre-crack in a thin plate [4.9]
4.9 CLOSURE
Problems dealing with plates of finite dimensions are more difficult to solve, because the edges
influence stress and displacement fields near the crack tip. However, approximate values of the
SIFs of some of these problems are determined through the Westergaard approach. In some
problems, the principle of superposition is helpful in determining the SIF.
For a given problem, the task of a designer is to determine the SIF somehow. For simple
problems, close form solutions available in reference books may be used. SIFs of more complex
problems may be available in handbooks on the stress intensity factor. The designer can also use
a computer package to evaluate the SIF. Once the SIF is known, he compares it with the critical
stress intensity factor, so as to ascertain that the crack is not likely to grow. If SIF exceeds the
critical SIF, the designer modifies his design—either to have a lower SIF (e.g., by changing the
crack configuration or reducing loads) or by changing material to have a higher critical SIF.
APPENDIX 4A
General Approach to Determine Stress and Displacement Fields
4A.1 MUSKHELISHVLI POTENTIALS
Solving the biharmonic equation through the Westergaard approach has its limitations, because
they are restricted to a few simple cases. For the general approach, Airy's Stress Function is
expressed in the form
Ú
F(z) = Re [z f (z ) + y ( z) dz]
SIF of More Complex Cases
85
where f (z) are y (z) two analytic complex functions. They are chosen appropriately to satisfy the
boundary conditions of the given problem.
Now, the goal is to express stress and displacement components in the terms of f (z) and y (z).
For the development of expressions, one may refer to the monograph of Muskhelishvli [4.3] or
other references [4.10]. The resulting stress field is given by
s11 + s22 = 4Re [f¢(z)] = 2[f¢(z) + f ¢ (z)]
s22 – s11 + 2is12 = 2[ z f ¢¢(z) + y ¢(z)]
(4.17a)
(4.17b)
where z = x1 – ix2 and prime on a function corresponds to differentiation with z. Unlike in Chapter 3,
the bar at the top of a complex number denotes complex conjugate. These two equations, in fact,
represent the three equations for the three unknowns—s 11, s 22 and s 12. The first equation deals
only with real numbers. The second equation is having both real and imaginary parts and thus
represents two equations, because real and imaginary parts may be equated separately. Functions
f (z) and y (z) are still unknown complex functions which, for a prescribed problem, are chosen to
satisfy the boundary conditions. Alternatively, approximate solutions may be obtained by
expressing each one of them in terms of a power series of z. Both approaches will be discussed
subsequently in this section.
4A.2 CENTRE-CRACKED PLATE
In Sec. 3.5.1, solution had been determined for a centre-cracked plate loaded by a biaxial field
because the Westergaard function is not available for a uniaxial field. However, the problem of a
uniaxial far field stress can be solved with the general approach. The expressions for f (z) and
y (z) are available, which satisfy all the near and far field boundary conditions. They are:
s
[2(z2 – a2)1/2 – z]
(4.18a)
4
s
y (z) =
[z – a2 (z2 – a2)– 1/2]
(4.18b)
2
with its origin at the centre of the crack [Fig.3.2(a)]. To show that they satisfy the boundary
conditions, we would be substituting f (z) and y (z) in Eqs (4.17). It is convenient to first evaluate
differentials of f (z) and y (z) as
f (z) =
f ¢(z) =
s
4
f ¢¢(z) = –
y ¢(z) =
s
2
È
˘
2z
- 1˙
Í 2
2 1/2
ÍÎ (z - a )
˙˚
˘
sÈ
a2
Í 2
2 3/2 ˙
2 ÎÍ (z - a ) ˚˙
È
˘
a2 z
+
1
Í
2
2 3/2 ˙
(z - a ) ˚˙
ÎÍ
To check the boundary conditions on the far field stress, we determine their values for |z|>> a as
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Elements of Fracture Mechanics
s
s
; f ¢¢(z) Æ 0; y ¢(z) Æ
4
2
Substituting in Eqs (4.17a and b), we have
s 11 + s 22 = s
f ¢(z) Æ
s 22 – s 11 + 2is 12 = s
leading to s11 = 0; s22 = s ; s 12 = 0
They satisfy the far field stress boundary conditions of the uniaxial stress. In order to check the
boundary conditions on traction free crack-faces, values of f ¢(z), f ¢¢(z) and y ¢(z) are determined
for x2 = 0 and –a < x1 < a as
f ¢(z) =
˘
2x1
s È
- 1˙
Í 2
2 1/2
4 ÎÍ (x1 - a )
˙˚
f ¢¢(z) = –
y ¢(z) =
˘
sÈ
a2
Í 2
2 3/2 ˙
2 ÍÎ (x1 - a ) ˙˚
˘
a2 x1
s È
Í1 + 2
˙
2 ÎÍ
(x1 - a2 )3/2 ˚˙
Substituting them in Eqs (4.17a and b), for – a < x1 < a, we have
s 11 + s 22 = – s
leading to
s 22 – s 11 + 2is 12 = s
s 22 = 0; and s12 = 0
Thus, the cracked surfaces are traction free.
For obtaining stress field in the vicinity of the crack tip, the origin may be moved to the crack
tip, as done in Sec. 3.5, with the transformation z = a + z0. It can be shown, by neglecting higher
terms of r/a, that the stress field is same as obtained earlier through Eqs (3.32). Better accuracy in
results can be achieved by retaining higher terms of r/a.
Many other problems can be solved by trying appropriate expressions for f and y for a given
problem. However, to arrive at such expressions, one relies on feel, experience or intuition.
4A.3 SOLUTION THROUGH P OWER S ERIES
The general way is to express f (z) and y (z) in terms of a power series. For the symmetric case of
a centre-cracked plate loaded in Mode I through a uniaxial stress, it has been shown [4.11] that
the f (z) and y (z) can be expressed as
f(z0) = A z 0l +1
(4.19a)
y (z0) = Bz 0l+ 1
(4.19b)
where the exponent l is real but A and B may be complex; z0 is measured from the crack tip. For
solving the problem, f (z) and y (z) will be substituted in Eqs (4.17a and b) and the boundary
conditions will be invoked. Boundary conditions on the cracked faces are:
SIF of More Complex Cases
87
s 22 = 0 for – a < x < a
(4.20a)
s 12 = 0 for – a < x < a
(4.20b)
Applying these boundary conditions, it can be shown that the problem be reduced to an
eigenvalue problem with multiple roots of l, each root having a solution. The overall solution would
then be the sum total of all the individual solutions. Equation (4.17a) is added to Eq. (4.17b) to have:
s22 + is 12 = 2Re[f ¢(z0)] + [ z0 f ¢¢(z0) + y ¢(z0)]
(4.21)
For carrying out the details, we differentiate Eqs (4.19a and b) to obtain:
f ¢(z0) = A(l + 1)zl0 = A(l + 1)rl (cos lq + i sin lq )
f ¢¢(z0) = Al(l +
= Al(l + 1)rl –1[(cos(l – 1)q + i sin(l – 1)q ]
y ¢(z0) = B(l +
(4.22a)
1)z 0l –1
1)z 0l
l
= B(l + 1)r (cos lq + i sinlq )
(4.22b)
(4.22c)
Substituting in Eq. (4.21), we have
s22 + is12 = 2 A(l + 1)r l cos lq + [r(cos q - i sin q )Al (l + 1)r l -1
¥ {cos (l - 1) q + i sin( l - 1) q } + B(l + 1) r l (cos lq + i sin lq )]
Equating real and imaginary parts and also by collecting terms, we obtain
s22 = (l + 1)r l [A{2cos lq + l cos q cos(l - 1) q + l sin q sin(l - 1) q } + B cos lq ]
= (l + 1) r l [A{2cos lq + l cos(l - 2) q } + B cos lq ]
(4.23a)
l
s 12 = (l + 1)r [ Al{cosq sin(l - 1) q - sin q cos(l - 1) q } + B sin lq ]
= (l + 1)r l [ Al sin(l - 2) q + B sin lq ]
(4.23b)
On crack faces (q = ±p) both s22 and s12 are zero, thus yielding
A{2cos lp + l cos(l - 2)p } + B cos lp = 0
(4.24a)
A l sin ( l - 2)p + B sin lp = 0
(4.24b)
These equations represent an eigenvalue problem with l representing the eigenvalues. Therefore,
for having nontrivial solution,
(2cos lp + l cos (l - 2)p )
l sin (l - 2)p
cos lp
=0
sin lp
leading to sin 2lp = 0. Consequently, for Mode I problems,
1 1 3 5
,
,
,
,...
2 2 2 2
Note that other negative possible values of l(–3/2, –5/2...) are nor admissible, because they
would give singular displacements at the vicinity of the crack tip. l = –1/2 is the dominant mode
and it gives singular stress and non-singular displacement. Other values of l do not give singular
stresses and therefore their contributions may be neglected if the analysis is carried out in a region
which is very close to the crack tip, i.e., for a very small value of r/a.
l=–
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Elements of Fracture Mechanics
The stress field near the crack tip corresponds to the dominant mode l = –1/2. Eigenvector
corresponding to this eigenvalue is determined through Eqs (4.24a and b), as
A = 2B
Substituting l = –1/2 and B = A/2 in Eqs (4.23a and b) and manipulating trigonometric functions,
we obtain
qÈ
q
A
3q ˘
cos Í1 + sin sin ˙
s22 =
(4.25a)
2Î
2
2 ˚
r
s12 =
q
q
A
3q
cos
sin cos
2
2
2
r
(4.25b)
Definition of KI [Eq. (3.4)] leads to
Aˆ
Ê
KI = Á 2p r
˜
Ë
r¯
rÆ0
KI
yielding,
A=
2p
Substituting A in Eqs (4.25a and b), we obtain
s22 =
KI
qÈ
q
3q ˘
cos Í1 + sin sin ˙
2Î
2
2˚
2p r
KI
q
q
3q
sin cos cos
2
2
2
2p r
For determining the value of s11, we substitute l = –1/2 in Eq. (4.22a) to evaluate f ¢(z0), which is
then substituted in Eq. (4.17) to yield
s12 =
s11 + s22 =
2K I
q
q
2A
cos =
cos
2
2
r
2p r
KI
qÈ
q
3q ˘
cos Í1 - sin sin ˙
2Î
2
2˚
2p r
The stress field thus obtained is same as that of Sec. 3.5.1. In order to predict more accurate
results, similar analysis may be carried out for the next higher term (l = 1/2) and be added to the
results already obtained for l = –1/2.
It is worth reviewing that the problem of a centre-cracked plate loaded in Mode I can been
solved in three different ways (Sections 3.5, 4A.2 and 4A.3), giving the same stress field for a
small value of distance r. The solution of Sec. 3.5 is always approximate because the problem was
solved for a biaxial filed. However, the solution of Secs 4A.2 and 4A.3 can be obtained with a high
accuracy if the contribution of higher terms is included in the solution.
A plate of finite dimensions is more difficult to solve because the stresses and displacements
must also satisfy the prescribed conditions at the boundary of the component. One can refer to
the work of Ramesh et al. [4.12] for the solution of a uniaxial field applied to a plate [4.12] of finite
dimensions. For these problems, Muskhelishvili potentials f(z) and y (z) are represented through
a series. The coefficients of the series can then be determined by matching the prescribed stresses
at the boundary [4.13, 4.14].
leading to,
s11 =
SIF of More Complex Cases
89
APPENDIX 4B
SIF of Some Important Cases
In this appendix, the stress intensity factor for some important cases is listed, which are useful
to (i) experimentalists for designing specimens and (ii) designers for determining the SIF for
critical components. For other cases, readers can refer to handbooks like “Stress Intensity Factor”
(2 volumes), edited by Y. Murakami, Pergamon Press, Oxford, 1987 [4.4].
1. Compact Tension (CT) Specimen, shown in Fig. 4.18(a) [4.15]
P
P
a
W
Fig. 4.18(a)
KI =
P
BW 1/2
f (a )
a
W
B = Plate thickness
a=
f (a) =
(2 + a ) (0.886 + 4.64a - 13.32a 2 + 14.72a 3 - 5.6a 4 )
(1 - a )3/2
2. Single-Edge-Notch-Bend (SENB) Specimen, shown in Fig. 4.18(b) [4.15]
P
W
a
S = 4W
Fig. 4.18(b)
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Elements of Fracture Mechanics
KI =
PS
f (a )
BW 3/2
a
W
B = Plate thickness
a=
f (a) =
(
)
3a 1/2 È1.99 - a (1 - a ) 2.15 - 3.93a + 2.7a 2 ˘
Î
˚
2 (1 + 2a ) (1 - a )
3/2
3. Centre-Cracked Plate under Uniform Tension, shown in Fig. 4.18(c) [4.5]
s
2a
2W
s
Fig. 4.18(c)
KI = s p a f (a )
a
a=
for 0 < a < 0.7
W
f (a) = 1.0 + 0.128a – 0.288a2 + 1.523a3
4. Single-Edge-Cracked Plate under Uniform Tension, shown in Fig. 4.18(d) [4.5]
s
a
W
s
Fig. 4.18(d)
SIF of More Complex Cases
KI = s p a f (a )
a
a=
for 0 < a < 0.6
W
f (a) = 1.12 – 0.23a + 10.55a 2 – 21.72a3 + 30.39a4
5. Double-Edge-Cracked Plate under Uniform Tension, shown in Fig. 4.18(e) [4.5]
s
a
a
2W
s
Fig. 4.18(e)
KI = s p a f (a )
a
a=
for 0 < a < 0.7
W
f (a) = 1.12 – 0.20a – 1.20a2 + 1.93a3
6. Strip with Edge-Crack under bending, shown in Fig. 4.18(f) [4.5]
M
M
W
a
Fig. 4.18(f)
KI =
6M
BW 2
p a f (a )
a
W
B = Plate thickness
f (a) = 1.12 – 1.40a + 7.33a2 – 13.083a3 + 14a4
a=
7. Circumferentially Cracked Round Bar under Tension, shown in Fig. 4.18(g) [4.6]
KI = s p a f ( b )
91
92
Elements of Fracture Mechanics
s
D
a
d
a
s
Fig. 4.18(g)
b=
d
D
Ê1 1 3
ˆ1
f ( b ) = Á + + b - 0.361b 2 + 7.33 b 3 ˜
Ëb 2 8
¯2
1
b
8. Circumferentially Cracked Round Bar under Torsion, shown in Fig. 4.18(h) [4.6]
Mt
D
a
d
a
Mt
Fig. 4.18(h)
KIII =
b=
16 Mt
p a f (b )
p D3
d
D
SIF of More Complex Cases
Ê 1
ˆ3
0.5 3
5
35 2
f( b ) = Á 2 +
+ +
b +
b + 0.21b 3 ˜
b
8 16
128
Ëb
¯8
1
b
9. An Oblique Edge Crack in a Semi-Infinite Plane, shown in Fig. 4.18(i) [4.4]
b
a
s
s
Fig. 4.18(i)
KI = F I s p a
KII = FII s p a
b
FI
FII
10°
20°
40°
60°
80°
0.162
0.305
0.625
0.920
1.098
0.174
0.271
0.365
0.306
0.119
10. Lapped Joint under Tension-Shear, shown in Fig. 4.18(j) [4.4]
Q
Bonded
area
2a
Q
2b
Fig. 4.18(j)
KI = Q
p Êa ˆ
FI Á , k ˜
a Ëb ¯
2b
93
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Elements of Fracture Mechanics
KII = Q
k=
p Êa ˆ
FII Á , k ˜
a Ëb ¯
(
)
3 1 -n 2 Q
2b E
Q : Load per unit width of plate
Êa ˆ
Ê aˆ
FI Á , k ˜ = b 1 Á ˜
Ë b¯
Ëb ¯
g1
2
Êa ˆ
Ê aˆ
FII Á , k ˜ =
+ b2 Á ˜
Ë b¯
Ëb ¯ p
k
g2
b1
0
0.1
0.2
0.3
0.4
0.5
g1
0.770
0.687
0.639
0.614
0.576
0.558
0.397
0.394
0.375
0.352
0.344
0.331
b2
g2
0.365
0.285
0.230
0.186
0.149
0.142
0.710
0.744
0.776
0.812
0.868
0.843
11. Arc-Shaped Tension (AT) Specimen, shown in Fig. 4.18(k) [4.15]
X
P
r1
a
W
r2
P
Fig. 4.18(k)
KI =
P
BW 1/ 2
a
a=
W
f (a )
SIF of More Complex Cases
95
B = Plate thickness
1/ 2
È
r1 ˆ ˘ È a
2 Ê
Í
¥
a
1
0.25
1
1
+
( ) Á r ˜˙ Í
Í
3/ 2
Ë
˙ Î (1 - a )
2 ¯˚
ÎÍ
[3.74 – 6.3a + 6.32a2 – 2.43a3]
12. Disc-Shaped Compact Tension (DCT) Specimen, shown in Fig. 4.18(l) [4.15]
È 3X
˘
f(a) = Í
+ 1.9 + 1.1a ˙
W
Î
˚
˘
˙ ¥
˙˚
1.35 W
P
P
a
W
Fig. 4.18(l)
KI =
P
BW 1/ 2
f (a )
a
W
B = Plate thickness
a=
f (a) =
(2 + a ) ¥ (0.76 + 4.8a - 11.58a 2 + 11.43a 3 - 4.08a 4 )
(1 - a )3/2
QUESTIONS
1. Why is the solution of collinear cracks in an infinitely long strip important?
2. Why is an edge crack more dangerous than a centre crack of the same length a?
3. In the case of a shallow elliptical edge crack, the SIF at the tip of the minor axis is higher
than the SIF at the tip of the major axis. This may sound contrary to our intuition. Can you
explain the results on physical grounds?
4. In a short fiber composite material, an embedded fiber does not act like a dangerous crack.
Explain it by assuming the fiber has a shape of an ellipse.
5. Why does material property E appear in the expression that relates GI with KI?
6. Energy release rate of various modes can be summed, whereas the SIF cannot be. Why?
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Elements of Fracture Mechanics
7. If a designer finds that a crack (e.g., slot) in a component gives KI > KIc , discuss the different
options available to avoid such a situation.
8. Why does KIc depend on the direction in a rolled plate?
9. Why do rolled or forged components have better toughness?
10. Name an application of bending and twisting a plate with a crack.
11. If we use a specimen with a large lateral dimension to find KIc , the accuracy of the
experimental results is high. But, in experiments to determine the SIF, specimens with
large lateral dimensions are not employed. Why not?
PROBLEMS
1. Determine value of KI for some cases listed in Appendix 4B for the following values:
(a) Centre-cracked plate under uniform tension; 2a = 60 mm, 2W = 140 mm, s = 150 MPa.
(b) Single-edge-notch three point bend (SENB) plate; a = 20 mm, S = 150 mm, W = 50 mm,
B = 25 mm, P = 20 kN.
(c) Single-edge-cracked plate under uniform tension; a = 30, W = 70 mm, s = 140 MPa.
(d) Strip with edge crack under bending; a = 25 mm, W = 60 mm, B = 20 mm, M = 2000 Nm.
2. A crack of 3 mm length is emanated from the surface of a 50 mm diameter hole in a large
plate (Fig. 4.19). Compute the maximum stress s that would not allow the crack to grow
if KIc = 55 MPa m , sys = 300 MPa, E = 207 GPa. [Hint: Treat as single-edge-crack under
tensile stress computed using Inglis solution Eq. (2.1)]
s
x2
50 mm
dia
x1
3 mm
s
Fig. 4.19 The figure of Problem 2
3. A steel flat (KIc = 90 MPa m , sys = 700 MPa) is being considered as a structural member
of a pedestrian bridge over a busy street. The flat is 100 mm high and 40 mm thick, and the
main load on the flat is a bending moment. If there is an edge crack on the tension side,
draw a curve for allowable bending moment vs crack length.
SIF of More Complex Cases
97
4. Determine KI for the pressure distribution acting on the cracked faces of an infinite plate
as shown in Fig. 4.20.
p0
p0
2a
Fig. 4.20 The figure of Problem 4
5. The distribution of load per unit area on the cracked faces of a centre crack in a large plate
is shown in Fig. 4.21. Determine the values of stress intensity factor.
s0
s0
2a
Fig. 4.21 The figure of Problem 5
6. Determine the values of KI for a center crack of length 2a in a large plate, subjected to a
triangular distribution of load per unit area, as shown in Fig. 4.22.
s0
s0
2a
s0
s0
Fig. 4.22 The figure of Problem 6
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Elements of Fracture Mechanics
7. Determine values of KI in some cases (given in the text or listed in Appendix 4B) for the
following values:
(a) Circumferentially cracked round bar under tension; a = 20 mm, D = 100 mm, P = 80 kN.
(b) Circumferentially cracked round bar in torsion; a = 10 mm, D = 100 mm, Mt = 40 kN.
(c) An oblique crack in a semi-infinite plate; a = 30 mm, b = 45°, s = 200 MPa.
(d) Embedded semi-elliptical crack; a = 15 mm, 2c = 90 mm, t = 40 mm, s = 160 MPa.
(e) Lapped bonded joint under tension shear; 2b = 8 mm, 2a = 40 mm, Width = 25 mm,
Q = 400 N/mm, v = 0.3.
8. Determine GI from the SIF of the center cracked infinite plate, subjected to a uniform
tensile stress s for both plane stress and plane strain. Check whether they agree with the
expression used in Sec. 2.5.
REFERENCES
4.1 Broek, D. (1982). Elementary Engineering Fracture Mechanics, Martinus Nijhoff Publishers,
The Hague.
4.2 Paris, P.C. and Sih, G.C. (1970). Stress Analysis of Cracks, ASTM STP 381, pp. 30–83.
4.3 Muskhelishvili, N.I. (1953). Some Basic Problems of the Mathematical Theory of Elasticity,
translated from Russian by Radok, J.R.M., Noordhoff, Leyden.
4.4 Murakami, Y. (1987). Stress Intensity Factors Handbook, Pergamon Press, Oxford.
4.5 Gdoutos, E.E. (2005). Fracture Mechanics—An Introduction, Springer, The Netherland.
4.6 Hellan, Kare (1985). Introduction to Fracture Mechanics, McGraw-Hill Book Company,
New York.
4.7 Irwin, G.R. (1962). The Crack Extension Force for a Part-Through Crack in a Plate,
Transactions ASME, Journal of Applied Mechanics, 29, pp. 651–654.
4.8 Irwin, G.R. (1957). Analysis of Stresses and Strains near the End of a Crack Traversing a
Plate, Journal of Applied Mechanics, 24, pp. 361–364.
4.9 Sih, G.C. (1977). Mechanics of Fracture, Plates and Shells with Cracks, Vol. 3, Noordhoff
International Publishing, Leyden.
4.10 Meguid, S.A. (1989). Engineering Fracture Mechanics, Elsevier Applied Science, London.
4.11 Kanninen, M.F., and Popelar, C.H. (1985). Advanced Fracture Mechanics, Oxford University
Press, New York.
4.12 Ramesh, K., Gupta, S., and Srivastava, A.K. (1996). Equivalence of Multi-parameter Stress
Field Equations in Fracture Mechanics, International Journal of Fracture, 79, pp. R37–R41.
4.13 Isada, N. (1970). On the Determination of Stress Intensity Factors for Some Common
Structural Problems, Engineering Fracture Mechanics, 2, pp. 61–79.
4.14 Williams, M.L. (1957). On the Stress Distribution at the Base of a Stationary Crack,
Transactions ASME, Journal of Applied Mechanics, 24, pp. 109–114.
4.15 Annual Book of ASTM Standards (1987). Metals Test Methods and Analytical Procedures,
ASTM Publications, Philadelphia, 03.01, pp. 552–57.
Chapter
Anelastic Deformation at the
Crack Tip
5
Out of His eye the sun was born.
Rig Veda
5.1 FURTHER INVESTIGATION AT
THE
CRACK TIP
The material in the vicinity of a crack tip is most affected. We therefore attempted to place a
magnifying lens at the crack tip in Chapter 4, and determined stress and displacement fields in its
vicinity. We, of course, took the simple route of assuming that the material remains elastic even at
the crack tip where the stresses are very large. The assumption is difficult to accept for most of the
engineering materials because they do not remain elastic at high stresses. Under real conditions,
the material in the vicinity of the crack tip deforms anelastically. In metallic components, the
material yields and flows to decrease the stress. Anelastic deformation also occurs in the vicinity
of the crack tip in components of other materials like plastics, composites, cardboards, etc.
Anelastic deformation has not been considered so far in the analysis of our models. Why is it
so? We have observed in the previous chapters that even elastic solution to a problem is quite
complex. The magnitude of complexity increases several folds if a plastic zone is included in the
mathematical modeling. Therefore, if the size of the anelastic zone is small in comparison to the
crack length, we assume, only for the sake of analysis, that the entire body remains elastic. Such
analysis, as mentioned in the earlier chapters, is known as Linear Elastic Fracture Mechanics
(LEFM). For a large plastic zone, more sophisticated analytical methods have been developed
which would be discussed in the subsequent chapters.
Now, we will investigate what happens at the crack tip when loads are applied on a component.
Similar to the behavior of a district magistrate who tries to diffuse a problem such as a communal
tussle, the material in the neighborhood of the crack tip tries to reduce the danger of high stresses.
In metals, plastic deformation occurs which is generally caused by nucleation and motion of
dislocations. The material flows in such a manner that high stresses are reduced dramatically.
Quite often the flow of material makes the crack tip blunt, which in turn decreases the magnitude
of stress components. Thus, many potential catastrophic failures are avoided just by the local
plastic deformation at the crack tip.
100
Elements of Fracture Mechanics
In highly extensible polymers, molecules may become oriented and can result in strain
crystallization [5.1]. This blunts the crack tip. Often, the morphology of a material is found to
change over a large area around the crack tip.
Those materials which are not able to release high stresses are usually found to have low
toughness. Diamond is a good example of such materials. Its inter-atomic bonds are so strong
that the material in the vicinity of a crack tip does not yield. Consequently, any plastic zone is not
formed at the crack tip and the material toughness is very low. According to a Chinese proverb,
trees which can bend under the blow of a high velocity cyclone survive better. To survive against
the danger of high stresses near the crack tip, the material should be able to deform anelastically.
Discussions in this chapter would be confined mostly to metals. Since plastic zone plays a vital
role in a fracture, we will investigate the shape of the plastic zone, yield planes, and the factors
which control the size of the plastic zone. A designer who tries to avoid fracture failure in a
component prefers that a large plastic zone is developed at the crack tip. This is contrary to the
likings of a conventional designer who tries hard to avoid yielding at any portion of a component.
In a way, the size of the plastic zone may be regarded as a parameter in representing the toughness
of a material. This approach has not been adopted so far owing to problems in predicting the size
accurately through analytic methods and also owing to difficulties involved in experimental
determination of the size of the plastic zone.
5.2 APPROXIMATE SHAPE AND SIZE OF THE PLASTIC ZONE
The stress field in the vicinity of a crack tip has been determined in Sec. 3.5, assuming the material
of the component remains elastic even at the crack tip. Knowing the stress field, we may invoke
one of the two commonly accepted yield criteria, Mises or Tresca, for evaluating the size and
shape of the plastic zone. Rigorous analysis becomes complex because two sets of constitutive
equations should be used, one for plastic deformation inside the plastic zone, and another for
elastic deformation outside the zone. The surface that separates the two zones is not known and
the rigorous analysis may require iterative solutions to many problems.
A rigorous analysis is complex as closed form solutions are not available for most problems.
We will determine only approximate solutions. The procedure adopted for finding the interface
between a plastic and an elastic zone is as follows:
Consider that one moves on a radial line from a far away place (where the stress field is definitely
elastic) towards the crack tip and a yield criterion is being applied continuously. As soon as the
material is found to yield, we mark that point as the interface between the elastic and plastic fields.
Similar consideration on other radial lines generates the shape and the size of the plastic zone. It is
worth noting that the material in the plastic zone no longer takes high stresses predicted by elastic
analysis. This, in fact, offsets the internal force balance, and therefore, the elastic field surrounding
the plastic zone should be corrected or re-evaluated for a more rigorous analysis. To incorporate
such a correction is quite complex and is beyond the scope of this book. The shape and size of the
plastic zone thus obtained through the procedure can only be accepted as approximate figures.
Before we apply a yield criterion, it is useful to determine the principal stresses. Considering a case
of Mode I, s33 is a principal stress because s13 = s23 = 0 for plane problems. We need to rotate axes in
x1 – x2 plane to determine the principal stresses s1 and s2. Figure 5.1 shows the Mohr circle, for which
s11, s22 and s12 are known through Eq. (3.32). The centre s0 and the radius R of the Mohr circle are
Anelastic Deformation at the Crack Tip
101
Shear
stress
x1
R
s0
s2
s22
s1
s11
s12
Normal
stress
x2
Fig. 5.1 Mohr circle to determine principal stresses
s0 =
s 11 + s 22
KI
q
=
cos
2
2
2p r
1/2
ÈÊ s - s ˆ 2
˘
KI
q
q
11
2 ˙
+ s 12
=
R = ÍÁ 22
cos sin
˜
ÍË
˙
2
2
2
2p r
¯
Î
˚
These equations determine principal stresses s1 and s2 as (defining the larger one as s1)
s1 = s 0 + R =
KI
qÈ
q˘
cos Í1 + sin ˙
2Î
2˚
2p r
(5.1a)
s2 = s 0 – R =
KI
qÈ
q˘
cos Í1 - sin ˙
2Î
2˚
2p r
(5.1b)
The third principal stress s 3 becomes,
s3 = 0
for plane stress
s 3 = n (s 1 + s 2) = 2n
KI
q
cos
2
2p r
(5.1c)
for plane strain
(5.1d)
5.2.1 Plastic Zone Shape for Plane Stress
Two widely used yield criteria, Mises and Tresca, are applied to Mode I, so as to determine the
plastic zone size for plane stress cases. All the three principal stresses and associated Mohr circles
are shown in Fig. 5.2(a). To ensure the yielding of the material, the Mises criterion states [5.2] that:
(s 1 – s 2)2 + (s 2 – s 3)2 + (s 3 – s 1)2 ≥ 2s ys2
(5.2)
where s ys is the yield stress. Substituting s 1, s 2 , and s 3, in the equation and using the symbol rpz
in place of r for the equality sign of the Mises criterion, we obtain
102
Elements of Fracture Mechanics
t
tmax
t
t
tmax
tmax
s3
s2
s3
s
s1
(a)
s2
s2
s1 s
s1
s3
(b)
s
(c)
Fig. 5.2 Mohr circles for (a) plane stress, (b) plane strain for small q, and
(c) plane strain for large q
2
2
KI2
qÈ
q Ê
qˆ
qˆ ˘
Ê
cos2 Í 4 sin 2 + Á 1 - sin ˜ + Á 1 + sin ˜ ˙ = 2s ys2
Ë
2p rpz
2 ÍÎ
2 Ë
2¯
2 ¯ ˙˚
which is simplified to
rpz =
1 KI2 Ê
3
ˆ
1 + sin 2 q + cos q ˜
2 Á
Ë
¯
4p s ys
2
The resulting shape of the plastic zone size is plotted in Fig. 5.3(a) in terms of non-dimensional
distance rpz/(K2I/(ps 2ys )). The shape of the plastic zone size is slightly different if the Tresca yield
criterion is invoked. In order to ensure yielding, the Tresca yield criterion [5.2] states that:
x2
x2
s1 - s 2
2
s1 - s 3
2
0.5
rpz
2
2
)
K I /(ps ys
rpz
2
)
K 2I /(ps ys
0.5
Plane
strain
x1
0.5
Plane
stress
Plane stress
Plane strain
(a)
(b)
Fig. 5.3 Plastic zone shape and size for (a) Mises yield criterion, and
(b) Tresca yield criterion
x1
Anelastic Deformation at the Crack Tip
tmax ≥
103
s ys
(5.3)
2
For plane stress, tmax is given by the biggest Mohr circle [Fig. 5.2(a)] which is between s 1 and
s 3. Thus, at r = rpz
s ys
s1 - 0
=
2
2
Substituting s1 from Eq. (5.1a), we obtain the approximate shape of the plastic zone as [Fig. 5.3(b)]
rpz =
KI2
2
2p s ys
È
qÊ
q ˆ˘
Ícos 2 ÁË 1 + sin 2 ˜¯ ˙
Î
˚
2
5.2.2 Plastic Zone Shape for Plane Strain
The third principal stress s 3 is no longer zero [Eq. (5.1d)] and, therefore, it influences the yielding
considerably. In fact, s 3 is also a tensile stress, creating conditions of triaxiality of the tensile
stresses. As a result, the plastic zone size is substantially smaller.
To find the shape of the plastic zone using the Mises yield criterion, s 1, s 2 and s 3 are substituted
in Eq. (5.2) and the resulting equation simplifies to
rpz =
KI2
2
4p s ys
2
È3
˘
2
ÍÎ 2 sin q + (1 - 2v ) (1 + cos q )˙˚
If the Tresca yield criterion is applied, we should first find out which is the largest Mohr circle.
Looking carefully into Eq. (5.1), we find s 1 is always larger than s 2 and s 3. Subtracting Eq. (2.1d)
from Eq. (2.1b), we obtain
KI
qÈ
q˘
cos Í(1 - 2v ) - sin ˙
2
2˚
2p r
Î
Note that for small q (£ 38.9° for n = 1/3), s 3 is the smallest principal stress and the yielding is
governed by s 1 and s 3 as shown in Fig. 5.2(b). Substituting s 1 and s 3 in Eq. (5.3), we have
s2 – s3 =
KI
qÊ
q
ˆ
cos Á 1 + sin - 2v ˜ = s ys
¯
2Ë
2
2p rpz
leading to
rpz =
q
2
2 Ê 1 - 2v + sin q ˆ
ÁË
˜
2
2¯
2p s ys
KI2 cos 2
For a large value of q (≥ 38.9° for n = 1/3), s 2 is the smallest principal stress and the yielding is
governed by s 1 and s 2 , as shown in Fig. 5.2(c). Then, rpz is obtained by using Eq. (5.3) as
rpz =
K I2
2
2p s ys
sin2q
Corresponding plastic zone shapes are shown in Fig. 5.3. As expected, the plastic zone size of
plane strain is much smaller than that of plane stress. In fact, the difference between the two
104
Elements of Fracture Mechanics
plastic zones is so large that it should convince a designer to choose, as much as possible, thin
plates or thin sections of a component.
It is worth noting here that the plastic zone size is proportional to the square of the stress
intensity factor and inversely proportional to the square of the yield stress. Whenever yield stress
of a metal is increased, say by an appropriate heat treatment, its plastic zone size decreases
considerably; this in turn makes the material more prone to crack growth. The increase in yield
stress may please a conventional designer, because he usually designs structural components
based on a yield criterion. But as far as the toughness of a material is concerned, the designer is
left with an inferior material. The designer should explore a happy compromise between yield
stress and toughness, while choosing a material and its heat treatment.
5.3 EFFECTIVE C RACK LENGTH
The appearance of the plastic zone at the tip does not allow its material to bear high stresses
predicted by the elastic analysis. Looking from a different angle, one can argue that the material
is soft in front of the crack tip and therefore the effective crack length is longer than the actual.
In fact, owing to the presence of the plastic zone, the stiffness of the component decreases or
the compliance increases. Consequently, the crack is equivalent to a length that is longer than
actual length. The size of the plastic zone in front of the crack tip determines the effective crack
length. Therefore, considerable efforts have been made by many investigators to determine the
plastic zone size accurately in front of the crack tip along the x1-axis.
5.3.1 Approximate Approach
One of the simplest (but highly approximate) expressions for the plastic zone size is found from
s22 vs x1 curve of Fig. 5.4 for Mode I. The length of the plastic zone r* along x1 direction is then
obtained from the relation
s22
ksys
x1
a
r*
Fig. 5.4 Approximate plastic zone size r*
Anelastic Deformation at the Crack Tip
s 22 = ks ys
105
(5.4)
where, k depends on whether the case is of plane stress or plane strain. Substituting s 22 =
KI/(2p r*)1/2and solving for r*, we obtain
r* =
KI2
(5.5)
2
2p k 2s ys
We now evaluate k for both plane stress and plane strain. Due to the symmetry, s 12 vanishes on
the x2 = 0 plane. Also, on this plane s22 = s 11. Thus, we have principal stresses as
s 1 = s 22
s 2 = s 22
s3 = 0
for plane stress
s 3 = n (s 1 + s 2) = 2n s 22
for plane strain
Substituting s 1, s 2 and s 3 in Eq. (5.2), and then comparing the resulting expression with Eq. (5.4),
we obtain
k=1
for plane stress
k = 1/(1 – 2n ) for plane strain
In fact, k = 3 for n = 1/3 in the case of plane strain. Substituting the value of k in Eq. (5.5), we
obtain
1
r* =
2p
r* =
ÊK ˆ
I
Á
˜
s
Ë ys ¯
1
18p
2
Ê KI ˆ
Á
˜
Ë s ys ¯
for plane stress, and
(5.6a)
for plane strain (n = 1/3)
(5.6b)
2
These results are highly approximate; better expressions are obtained through the Irwin plastic
zone correction, or the Dugdale approach.
5.3.2 The Irwin Plastic Zone Correction
Irwin [5.3] suggested a vastly improved expression for the plastic zone size along the x1-axis
through a model which accounts for the absence of high stresses within the yield zone. Consider
a case of plane stress with k = 1. As shown in Fig. 5.5, the plastic zone size beyond the tip T of the
actual crack is TL along x1-axis. The tip of the effective crack is located somewhere inside the
plastic zone. There should be an appropriate criterion to find the length of the effective crack.
Note that if the tip of the effective crack is at point M, then the material beyond the tip will not be
able to sustain s 22 > s ys. Moreover, the crack-faces of the effective crack are not actually separated
between the points T and M and therefore tensile stress equal to s ys acts on the length d for an
elastic-perfectly plastic material. Irwin proposed that d is so chosen such that the load not taken
beyond the point M, given by the area Pc on length l, is equal to the load sustained on length d ,
given by area PA. Then, the load not sustained on length l becomes
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Elements of Fracture Mechanics
Èl
˘
Pc = B Í s 22 dx1 - s ys l ˙
Í
˙
Î0
˚
Ú
s22
Pc
sys
PA
T
M
x1
L
d
a
l
Fig. 5.5 Irwin plastic zone correction
where, B is the plate thickness. The load sustained (PA ) on length d is
PA = Bs ys d
Irwin's correction (PA = Pc) leads to
s ys d =
l
l
0
0
Ú s 22 dx1 - s ysl = Ú
KI
dx1 - s ysl
2p x1
(5.7)
where, KI is based on the effective crack length (a + d ). l is determined by noting that at x1 = l ,
s 22 is equal to s ys for plane stress (k = 1), i.e.,
KI
s ys =
2p l
Rearranging, we have
KI = (2p l)1/2s ys
Substituting in Eq. (5.7), we have
l
s ysd =
Ú
0
On solving, we obtain
d =l
s ys ( 2p l )
1/2
2 p x1
dx1 - s ys l
Anelastic Deformation at the Crack Tip
107
The overall plastic zone size becomes
rp = 2l =
KI2
(5.8)
2
p s ys
The effective crack length aeff is given by
KI2
aeff = a + d = a + l = a +
2
2p s ys
(5.9)
Since KI is based on the effective crack length, aeff is still unknown. If the plastic zone is small in
comparison to the crack length, KI may be assumed to be equal to the SIF of crack length a.
However, KI can also be obtained in a closed form for the case of an infinite plate; the corrected
SIF is expressed as
KI = s ÈÎp ( a + l ) ˘˚
= sp
1/2
1/2
È
KI2 ˘
Ía +
˙
2
2p s ys
ÍÎ
˙˚
1/2
On solving, we obtain
KI =
s p a
(
)
1/2
2˘
1
È
s
s
1
/
ys
ÍÎ
˙˚
2
(5.10)
For a finite size specimen,
1/2 Ê a + l ˆ
KI = s ÈÎp ( a + l ) ˘˚ f Á
Ë W ˜¯
or
1/2
ÊÊ
ˆ
È
KI2 ˘
KI2 ˆ
˙
KI = s p Í a +
(5.11)
+
f
a
W
Á
˜
Á
˜
2
2
ÁË Ë
˜¯
2p s ys
2p s ys
ÍÎ
˙˚
¯
For certain geometries, function f is simple (e.g., f = 1 for a centre-cracked infinite plate and
f = 1.12 for an edge-cracked infinite plate) and KI can easily be obtained from the above equation.
Otherwise, it can also be determined by an iterative procedure. For example, KI on right hand
side is taken based on the actual crack length a in the first round of iteration. The evaluated value
KI is then fed on the right hand side in the second round. The iterative procedure is repeated until
two successive values of KI are within a small percentage difference.
Irwin's correction to the plane strain case is useful to determine the plastic zone size. Due to
the plastic deformation the crack tip becomes rounded. Since the rounded tip acts as a free surface,
s 11 is released to zero. The effect of the release of s 11 is felt for some distance on the xl-axis
beyond the crack tip. Irwin [5.4] found that k is no longer 3 but is closer to 2 2 which may be
2
taken as 3 . Then, the plastic zone size for the plane strain becomes (1/3p)K I2/s ys
. For an
experimental determination of KIc of material, plane strain conditions are assured by taking plate
1/2
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Elements of Fracture Mechanics
thickness to be much thicker than the plastic zone size. In fact, it is chosen to be more than 25
times of the plastic zone size and therefore, the ASME codes require
B ≥ 2.5
K I2
2
s ys
The details of the experimental procedure to determine the critical SIF are presented in Chapter
8. To have an idea, the estimated plastic zone size (rp) is listed in Table 5.1 for some commonly
used materials for plane strain, corresponding to KIc.
TABLE 5.1 Estimated plastic zone size (rp) at the critical stage of a crack for some
common materials for plane strain in Mode I
Material
rp (mm)
Nuclear Reactor Steel
Alloy Steel* ( s ys = 1070 MPa)
High Strength Aluminum (7075-T651)
Titanium Alloy (Ti-6Al-4V)
Perspex (Plexiglass)
93
1.8
0.8
1.3
0.1
* 40 Ni2Cr1Mo28 (IS)/EN 24 (UK)/4340 (USA)
5.3.3 Plastic Zone Size through the Dugdale Approach
Dugdale [5.5] determined the plastic zone size through a different approach. He considered the
effective crack to be of length a + r where r is plastic zone size [Fig. 5.6(a)].
s
sys
sys
sys
sys
r
a
a
r
sys
ds
sys
s
s
ds
s
(a)
(b)
Fig. 5.6 (a) Plastic zone size through Dugdale’s approach and (b) nullifying the
singularity using Green’s function approach
In the Dugdale approach, singularity at the tip of the effective crack is nullified by a uniform
pressure equal to yield stress s ys (for plane stress cases with k = 1) on the portion r of the crack, as
shown in the figure. In fact, to determine r we employ the criterion that r is the length on which
pressure s ys exactly nullifies the singularity. If K r is the singularity due to the pressure s ys and Ks
due to external load on the effective crack, we have
Anelastic Deformation at the Crack Tip
Ks + Kr = 0
109
(5.12)
Ks is simple to evaluate and is given by
Ks = s p ( a + r )
To determine Kr , Green's function is used (as discussed in Sec. 4.1.1). Application of Eq. (4.1) to
Fig. 5.6(a) gives
dKr = –
2 s ys ds p (a + r )
p ( a + r )2 - s 2
Integrating, we have
Kr = –
2s ys p (a + r )
p
a+ r
Ú
a
1
( a + r )2 - s 2
ds
2s ys p (a + r ) È -1 a ˘
Í cos
˙
p
a+ r˚
Î
Substituting Kr and Ks in Eq. (5.12), we have
=–
s–
2s ys
p
cos -1
a
=0
a+r
which is simplified to
È ps ˘
a
= cos Í
˙
a+r
ÍÎ 2s ys ˙˚
An approximate, but simpler relation may be obtained for cases s << sys and r << a as
1–
giving,
r
p 2s 2
=1–
2
a
8s ys
r=
ps 2p a
2
8s ys
=
p KI2
(5.13)
2
8s ys
We note that the plastic zone size of Dugdale's model is close to that of Irwin's correction [i.e.,
K I2/(ps ys2 )] for s << sys.
Example 5.1 A large plate of 5 mm thickness, made of medium carbon steel (s ys = 350 MPa) with
a through-the-thickness centre-crack of 2a = 40 mm length, is subjected to a stress of 150 MPa. For
Mode I loading, determine the effective crack length using Irwin's correction.
Solution: We start by assuming that the plate is loaded in plane stress.
KI = s
rP =
p a = (150 MPa)
KI2
2
ps ys
=
p ¥ (0.020 m) = 37.6 MPa m
(37.6 MPa m )2 = 3.67 mm
p (350 MPa)2
110
Elements of Fracture Mechanics
Since rP is not negligible in comparison to plate thickness, our assumption of plane stress
conditions is justified. Invoking Eq. (5.10), we have:
KI =
s pa
2 ˘1/2
È
Ê
ˆ
Í1 - 1 Á s ˜ ˙
Í
2 Ë s ys ¯ ˙
Î
˚
1/2
= (150 MPa) p ¥ (0.020 m)
È
Í1 Í
Î
1/2
2
1 Ê 150 MPa ˆ ˘
˙
2 ÁË 350 MPa ˜¯ ˙
˚
= 39.46 MPa m
The plastic zone size is given by
rp =
KI2
2
ps ys
=
(39.46 MPa)2
p ¥ (350 MPa)2
= 0.004046 m = 4.046 mm
And then the effective crack length becomes:
rp
aeff = a +
2
= 20 mm + 2.023 mm = 22.023 mm
Example 5.2 A steel plate (sys = 350 MPa) of width 80 mm and thickness 5 mm has a centrecrack 2a = 40 mm length. If the far field stress is 150 MPa, determine the SIF and the length of the
effective crack, using Irwin's correction.
Solution: We first check whether the plate is loaded in plane stress by finding rp based on the
actual crack length. Referring to Appendix 4B, we have
KI = s
p a f (a );
a = a/W =
20 mm
= 0.5
40 mm
f (a) = 1.0 + 0.128a – 0.288a 2 + 1.523a 3
= 1.182
KI = (150 MPa m ) p ¥ 0.02m ¥ 1.182 = 44.44 MPa m
rp =
KI2
2
ps ys
2
=
1 Ê 44.44 MPa m ˆ
= 5.13 mm ª Plate thickness
p ÁË 350 MPa ˜¯
Therefore, plane stress conditions do exist. Invoking Eq. (5.11), we have
1/2
È Ê
KI2 ˆ ˘
˙
KI = s Íp Á a +
2 ˜
˜¯ ˙
Í ÁË
2ps ys
Î
˚
where,
¥ f (a ¢)
(5.14)
Ê
KI2 ˆ
a ¢ = Áa +
W
2 ˜
2ps ys
Ë
¯
This problem will now be solved through iteration. The general iterative procedure is that we
guess a value of KI and substitute it on the RHS of Eq. (5.14) to evaluate the variable KI of LHS
Anelastic Deformation at the Crack Tip
111
and then compare the two values. In the second round of iteration, we guess another value of KI
on the RHS. The process continues till the two values converge. In this problem, we start the first
round of iteration with KI = 44.44 MPa m and substitute it in the RHS of Eq. (5.14) to have
a ¢ = 0.5641,
f(a ¢) = 1.2539
KI = 50.08 MPa m
Since this KI is quite different from the 44.44, we go for the second round of iteration by feeding KI
as 50.08 into RHS of Eq. (5.18) to have
a ¢ = 0.5815,
f(a ¢) = 1.2765
KI = 51.758 MPa m
The third round gives
a ¢ = 0.587,
f(a ¢) = 1.2839
KI = 52.306 MPa m
and the fourth round determines
a ¢ = 0.5889,
f(a ¢) = 1.2865
KI = 52.48 MPa m
Since KI is quite close to the value of the third round, we stop the iterative process. The plastic
zone size is then given by
rp =
KI2
2
ps ys
=
(52.48 MPa m )2
p ¥ (350 MPa)2
= 0.00716 m = 7.16 mm
and the effective crack length becomes,
aeff = a +
rp
2
= 0.02 m + 0.00358 m = 23.58 mm
5.4 EFFECT OF P LATE T HICKNESS
So far in this book, we have been using the terms plane stress and plane strain without having
spelled out any formal quantitative criterion, which would help us identify whether a given case
is of plane stress or plane strain. Such a criterion could not be given earlier because the parameter
that differentiates between plane stress and plane strain is the size of the plastic zone. Only after
quantitative models are developed, we can estimate the plastic zone.
In a thin plate, plane stress components s 33, s 31 and s 32 and are able to relax to zero and the
material deforms easily within the plastic zone. In the vicinity of the crack tip, both free surfaces
of the plate move in forming a depression on each side, owing to in-plane tensile stresses and
Poisson's effect. On the contrary, the material in a thick plate is constrained, giving rise to tensile
stress s 33.
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Elements of Fracture Mechanics
There must be a limit on plate thickness under which a material is able to flow easily and the
plate is deformed under plane stress. Similarly, there must be a limit on plate thickness over
which the material is taken to be deformed under plane strain. Between the upper and lower
limits of the plate thickness, the cases are known to have transitional behaviour, i.e., near a free
surface, the material flows easily and deforms in plane stress and, in the interior, the material is
constrained and is subjected to plane strain.
Now, the method to set the limits on the thickness for plane stress and plane strain will be
explored. These limits are decided mainly on the basis of the long experiences of many
investigators. It has been found that if the plastic zone size is about the thickness of the plate, the
material within the plastic zone relieves 'out of plane stresses' and deforms easily. Therefore, for
a plate with its thickness less than or equal to the size of the plastic zone, the crack is loaded in
plane stress. Figure 5.7(a) shows the case of plane stress with a section through the plastic zone.
s
s
s
s
(a)
s
(b)
s
(c)
Fig. 5.7 Plastic zone size for (a) plane stress, (b) transitional case, and (c) plane strain
The thick plate [Fig. 5.7(c)] corresponds to plane strain, showing a smaller plastic zone. Even
in this case, some effect of the free surface exists where the plastic zone is larger. However, the
thick region of plane strain dominates and the surface effects can be neglected. A plate having
thickness greater or equal to 2.5 K2Ic/s 2ys is regarded as a case of plane strain. In transitional cases
[Fig. 5.7(b)], the interior portion of the plate and the regions near its both surfaces have mixed
effects on the plastic zone.
It is evident from Fig. 5.7 that the critical SIF of a plate depends upon its thickness. The
nature of the critical SIF dependence on the plate thickness B is shown in Fig. 5.8 qualitatively.
For B ≥ 2.5 K2Ic/s 2ys , the critical SIF remains constant and then we can regard the critical stress
intensity factor as the material property. The value of K Ic of commonly used materials is
available. Representative KIc of some materials are presented in Table 4.2.
Anelastic Deformation at the Crack Tip
Transitional
Plane
strain
Critical SIF
Plane
stress
113
KIc
B0
Plate thickness
B
Fig. 5.8 Variation of critical SIF with plate thickness
For B < 2.5 K2Ic/s 2ys critical stress intensity factor depends on the thickness B. The relation
between critical SIF and thickness should be provided to designers. However, this is not usually
done, probably because experimental tests are expensive for tough materials. Users like aerospacecompanies generate their own data base of critical SIF of plane stress or transitional cases, for
commonly used thicknesses.
For plate thickness B0 (Fig. 5.8), the plastic zone size is approximately equal to the thickness of
the plate thickness and is considered to be the case of pure plane stress. For thinner plates,
experimental determination of the critical SIF is difficult, because the plate tries to buckle close to
the cracked surfaces and the specimen is then loaded in Mode III also.
Fractured planes of a plane stress case are different from those of a plane strain case. In case of
Mode I, s 33 is always zero if conditions of plane stress prevail. Also on plane x2 = 0, it is noted that
s 12 = 0 and s 22 = s 11. However, even at a point slightly away from the plane (q is not equal to
zero)s 22 is greater than s 11. The maximum shear stress is dictated by s 22 and s 33 and therefore, a
fracture is likely to occur on one of the two planes inclined at ± 45°, as shown in Fig 5.9(a). Note
that the fractured surface is inclined to the free surfaces of the plate as shown in Fig. 5.9(b).
Under plane strain conditions in Mode I, yield planes are more complex. The crack growth is a
combined effect of yielding due to the dislocation motion and the generation and nucleation of
voids in front of the crack tip. In metals, the nucleation and the growth of voids are facilitated by
the triaxiality of tensile stresses near the crack tip in case of plane strain. The voids coalesce (grow
and join each other) during the crack growth. It has been observed that in plane strain cases, a
crack usually grows on the plane of the original crack. Figure 5.10 shows the self similar growth
on the crack-plane. This fact is exploited in designing experiments, so as to determine the critical
SIF of a material.
114
Elements of Fracture Mechanics
x2
s
x1
x3
Lip
45°
45°
45°
Crack growth
s
(a)
(b)
Fig. 5.9 Yield planes of plane stress, and (b) a cracked face
s
Crack
growth
s
Fig. 5.10 Crack growth of plane strain cases
For the transitional behavior of a crack, the fracture is of mixed kinds (Fig. 5.11), flat in the
interior, dominated by plane strain and slant close to the free surface. In fact, by looking at a
fractured surface, one can conclude whether the fracture occurred prominently under plane stress
or plane strain.
Anelastic Deformation at the Crack Tip
115
Crack growth
(a)
(b)
Fig. 5.11 Cracked faces of transitional cases
Cutting a material by conventional machines (lathe, milling, shaper and drill press) is a fracture
phenomenon and the edges of a cut surface are found to have lips generally known as burrs.
These burrs are regarded as a nuisance in a production shop and are removed through specially
designed deburring machines.
5.5 CLOSURE
The shape of the plastic zone has been obtained through a quantitative, but approximate method.
This helps us in understanding the difference between plane stress and plane strain conditions.
The critical stress intensity factor depends, in general, on the plate thickness. However, it does
not depend on plate thickness, if the plates are thick enough to be loaded in plane strain. Thus,
the critical stress intensity factor of plane strain (KIc) is treated as the property of a material.
QUESTIONS
1. The shape of the plastic zone, as determined in this chapter, is approximate. Why?
2. In comparison to a plane strain case, a plane stress loading gives much larger plastic zone
for the same SIF? Why?
3. Looking at a fractured surface, can you distinguish whether the loading was in plane
stress, plane strain or transitional?
4. Show the yield planes of plane stress cases through a clear diagram.
5. The plastic zone size obtained through Irwin's model is quite large in comparison to the
plastic zone (r*), obtained through the yield criterion applied to the elastic field. Why?
6. Can the effective crack length be used as a suitable parameter for formulating elastic plastic
fracture mechanics?
7. Why is the fracture plane of plane strain case normal to the free surface and in the plane of
the original crack surface for Mode I loading?
8. Why is burr developed during machining?
9. Does fracture mechanics recommend the enhancement of the yield stress of an alloy
through a heat treatment? Justify your answer.
116
Elements of Fracture Mechanics
PROBLEMS
1. Show that the approximate plastic zone shape around the crack tip of a Mode II crack in
an infinite plate is given by:
1
rp(q ) =
8p
2
Ê KII ˆ
2
Á
˜ (14 – 2 cos q – 9 sin q )
s
Ë ys ¯
for plane stress employing Mises yield criterion. Plot the shape of the plastic zone.
2. Apply the Mises yield criterion to show that the approximate plastic zone shape around
the crack tip of a Mode II crack in an infinite plate for plane strain loading is given by
rp(q ) =
1
8p
2
ÊK ˆ
2
2
II
Á
˜ [12 + 2 (1 – 2 v) (1 – cos q ) – 9 sin q )]
s
Ë ys ¯
Plot the shape of the plastic zone.
3. Invoke the Mises yield criterion to show that the approximate plastic zone around the tip
of a Mode III crack is a circle of radius
3 Ê KIII ˆ
rp =
Á
˜
2p Ë s ys ¯
2
4. In a case of a mixed mode with loading in plane stress, find the approximate shape of the
plastic zone, if KI = KII = K, and plot the shape represented by rp/(K2/p s 2ys) vs q. Use of a
computer is recommended.
5. Determine the plastic zone size rp/a of an edge crack of length a in a thin infinite plate,
loaded in Mode I with a far field stress s using Irwin's correction.
6. Apply Irwin's correction to determine the SIF, the length of the effective crack and the
plastic zone size for an edge crack of 15 mm length in a plate 80 mm width. The thickness
of the plate is 5 mm and the far field stress is 150 MPa (s ys = 350 MPa).
7. A thin plate with an edge crack of 35 mm length is loaded in Mode I with a far field stress
of 300 MPa. If the yield stress of the material is 900 MPa and the material is idealized as
elastic-perfectly plastic, determine the plastic zone size on x1-axis using Irwin's correction.
What is the length of plastic zone size if the Dugdale Approach is applied?
REFERENCES
5.1 Atkins, A.G. and Mai, Y.W. (1985). Elastic and Plastic Fracture: Metals, Polymers Ceramics,
Composites, Biological Materials, Ellis Horwood, Chichester.
5.2 Srinath, L.S. (1980). Advanced Mechanics of Solids, Tata McGraw-Hill Publishing Co. Ltd,
New Delhi.
5.3 Irwin, G.R. (1958). Fracture, Handbuch der Physik, Springer, Berlin, Vol. VI, pp. 551–590.
5.4 Irwin, G.R. (1968). Linear Fracture Mechanics, Fracture Transition and Fracture Control,
Engineering Fracture Mechanics, I. pp. 241–257.
Anelastic Deformation at the Crack Tip
117
5.5 Dugdale, D.S. (1960). Yielding and Steel Sheets Containing Slits, Journal of Mechanics and
Physics of Solids, 8, pp. 100–108.
5.6 Anderson, T.L. (2004). Fracture Mechanics: Fundamentals and Applications, CRC, Press-Book.
5.7 Sanford, R.J. (2003). Principles of Fracture Mechanics, Prentice Hall, Upper Saddle River.
5.8 Janssen, M., Zuidema, J. & Wanhill, R.J.H. (2004). Fracture Mechanics, Spon Press, Abingdon.
5.9 Ramesh, K. (2007). e-Book on Engineering Fracture Mechanics, IIT Madras,
URL: http//apm.iitm.ac.in/smlab/kramesh/book_4.htm
Chapter
6
J-Integral
Siddharth said: “Is it not true my friend, that the river has very many voices? Has it not
the voice of a king, of a warrior, of a bull, of a might bird, of a pregnant woman and a
singing man, and a thousand other voices?”
Hermann Hesse
6.1 RELEVANCE
AND
SCOPE
A large plastic zone at the crack tip makes a material tough. However, in the early development of
modern fracture mechanics, existence of the plastic zone was ignored in any analysis, because
practical methods were not developed to account for the elastic-plastic behavior within the plastic
zone. The LEFM assumes that the entire body, including the material very close to the crack tip,
follows linear elastic relations.
The stress-strain behavior of an elastic-plastic deformation is not simple, as it is well known to be
non-linear. With the availability of inexpensive computers, non-linear stress-strain behavior can now be
accounted for. But, it is difficult to incorporate the non-uniqueness of the elastic-plastic behavior. That
is, the loading behavior of an elastic-plastic material differs from the unloading behavior, as shown in
Fig. 6.1. For a given value of strain (e.g., at e = e1), there are always two stress values and therefore, one
has to keep track of the specimen being loaded or unloaded. An analogy can be drawn with
interaction with a person who is simple and straight forward like a linear elastic material. Non-linear
elastic material corresponds to a man who is honest but rude. But elastic-plastic material is like
dealing with a man who is not only rude but speaks truth sometimes and lies other times.
For many metals, the stress-strain relations get further complicated due to the Bauschinger’s
effect [6.1], which shows that the yield stress in unloading is different from that in loading. Such
problems will not be elaborated here. But we realize that solving problems of an elastic-plastic
material is quite difficult, even for a regular component with no voids or cuts. The magnitude of
complexity increases further when the problem is solved in the vicinity of a crack tip.
Elastic-plastic behavior makes the analysis of fracture mechanics complex, because this implies
the existence of two zones—the elastic-plastic zone, near the crack tip and the elastic zone
surrounding it. The interface between the two zones is also not known and could be determined
only by a complex analysis.
J-Integral
119
B
s
Non-linear elastic
A
D
Elastic-plastic
Linear
elastic
e1
e
Fig. 6.1 Non-linear elastic and elastic-plastic material behavior
Because of these complex problems faced in any elastic-plastic analysis, it has, so far, not been
possible to tackle the problem directly. In fact, we look for clever solutions which are improved
over those of the LEFM, but are not too difficult for designers in field applications. The J-Integral is
one such method which, in spite of some reservations, is the most accepted method to designers
and researchers of elastic-plastic fracture mechanics (EPFM). In this chapter, we will discuss
various aspects of the J-Integral.
6.2 DEFINITION
OF THE
J-INTEGRAL
Like other parameters (G and K), the J-Integral is also a parameter to characterize a crack. In fact, G
is a special case of the J-Integral, i.e., G is usually applied only to linear elastic materials, whereas
the J-Integral is not only applicable to linear and non-linear elastic materials, but is considered to
be very useful to characterize materials, exhibiting elastic-plastic behavior near the crack tip.
At first sight, the J-Integral looks like a strange term, but as we go along and develop it gradually,
it would not look odd. For plane problems, consider a path G around a crack tip (Fig. 6.2) which
starts from any point of a crack face and ends on any point on the other crack face. The path can be
chosen arbitrarily within the material of the component. It may be smooth or may have corners,
but should be continuous. J-Integral was first applied to fracture mechanics by Rice [6.2] in 1968 for
plane problems. It is defined as:
J =
Ú
G
where,
Ê
∂ui ˆ
ÁË W dx2 - Ti ∂x ds˜¯
1
(6.1)
W = Ú s ij de ij
The Einstein’s summation convention is followed in defining the above expressions. For two
dimensional plane cases W may be expressed in full form as:
120
Elements of Fracture Mechanics
x2
G
Ti
ds
x1
ni
Fig. 6.2 Path G around the crack tip with outward normal ni and traction Ti
W =
e11
e12
e 21
Ú
s 11 de11 + Ú s 12 de12 +
Ú
0
0
e 22
s 21 de 21 +
0
Ú s 22 de 22
0
Summing the middle terms, we have:
e11
W =
Ú
0
e12
e 22
s 11 de11 + 2 Ú s 12 de12 + Ú s 22 de 22
0
0
Also, W which is strain energy per unit volume is a point function, i.e., it varies from point to point
within the body of the component. Other parameters of Eq. (6.1) are:
Ti = traction vector at a point on the path G,
ui = displacement vector at a point on the path.
In the expanded form, the second term of Eq. (6.1) for plane cases is:
Ú
G
Ê ∂u1
∂u2 ˆ
ÁË T1 ∂x + T2 ∂x ˜¯ ds
1
1
Traction Ti at a point on path G is expressed through the well-known relation Ti = s ij n j (i.e.,
T1 = s 11 n1 + s 12 n2 and T2 = s 21 n1 + s 22 n2 ). Thus, by knowing the stress field and the direction of the
normal at a point of path G, we can determine Ti. Later in this chapter, we will prove the following
two important features of J-Integral:
(i) The J-Integral is path independent, i.e., G can be chosen arbitrarily within the body of a
component.
(ii) For linear elastic bodies, the J-Integral represents energy release rate and is same as G.
Now, we will take up an example to show that the above mentioned features of the J-Integral are
applicable to a practical case of the double beam cantilever solved in Sec. 2.7.
J-Integral
121
Example 6.1 Determine the J-Integral for a double cantilever beam (DCB) specimen, if each
cantilever is pulled by a distributed load P, as shown in Fig. 6.3.
x2
F
E
P
h
H
B
x1
a
P
h
G
D
C
Fig. 6.3 Path G coinciding with outer edges of the DCB specimen
Solution: The chosen path G is BCDEFH and it coincides with the body contour. Segments CD and
EF of the path G do not contribute to the J-Integral, because dx2 is negligible and Ti = 0. The
contribution of the segment DE towards the J-Integral is negligible, because stresses are very small,
which in turn, make W and Ti negligible. On segments BC and FH, W is negligible, (no contribution
of bending stress owing to zero bending moment and negligible contribution of shear stress) and
the only non-trivial term is:
h
J = - 2Ú T2
0
∂u2
dx2
∂x1
(6.2)
where, the factor 2 accounts for the contribution from both cantilevers. Deflection u2 of a cantilever
at a distance x1 is commonly evaluated by equating the bending moment with EI
∂2 u2
∂x12
=
∂ 2 u2
, leading to:
∂x12
Px1
EI
Integrating it for the upper cantilever with the boundary condition ∂u2 / ∂x1 = 0 at x1 = a , we obtain:
Px12 Pa2
∂u2
=
∂x1
2 EI 2 EI
Evaluating it at x1 = 0 and using the relation I = Bh3/12, we have:
6 Pa2
∂u2
=∂x1
EBh3
Substituting in Eq. (6.2), we have:
J =
12 Pa2
EBh3
h
Ú T2 dx2
0
Elements of Fracture Mechanics
122
h
But B Ú T2 ds = P on face FH, leading to:
0
JI =
12 P 2 a2
EB 2 h 3
(6.3)
This expression is same as that of G1, obtained in Sec. 2.7 [Eq. (2.20)]. One may note here that the
definition of the J-Integral is more versatile than the definition of G, because the J-Integral can cut
out a portion of the component at the will of an investigator and characterize the crack fully.
Although the technique to determine J is quite different from that of G, both lead to the same result
for linear elastic materials in the end. Of course, the parameter G is limited only to LEFM, whereas
the J-Integral encompasses a much bigger domain, because it can deal with both non-linear and
elastic-plastic materials.
6.3 PATH INDEPENDENCE
The J-Integral is carried out along an arbitrary path G, which starts from one crack face and ends up
on the other crack face, while going around the crack tip. It would be proved, in this section, that J
is path independent for both linear and non-linear elastic materials. We first consider an area A of
a plate not containing any cut or void and we surround it by a closed path S, as shown in Fig. 6.4(a).
We will show that the value of integral I, containing the same integrand as that of the J-Integral, but
evaluated along the closed path S, vanishes, i.e.
I=
È
∂u
˘
Ú ÍÎW dx2 - Ti ∂x1i ds˙˚ Æ 0
(6.4)
S
A
n
S
S
ds
dx2
n2
q
q
dx1
s=0
(a)
n1
dx2 = cos q ds = n1 ds
(b)
Fig. 6.4 (a) Area A without any cut or void and enclosed by contour S, and
(b) the relation between n1 and dx2
Substituting dx2 = n1ds [Fig. 6.4(b)] and Ti = s ij n j in the above equation, we have:
I=
È
Ú ÍÎW n
1
S
- s ij n j
∂ui ˘
˙ ds
∂x1 ˚
J-Integral
123
The integral will be converted from a line integral to an area integral using the Divergence
Theorem, which is stated in general form as:
Ú Mn ds
i
=
∂M
Ú ∂xi dA = Ú Mi dA
A
S
A
where M is any variable. Invoking the Divergence Theorem, we have:
I=
Ú
È ∂W Ê ∂u ˆ ˘
Í
- s ij i ˙ dA
ÍÎ ∂x1 ÁË ∂x1 ˜¯ , j ˙˚
∫
 ∂W ∂ε ij
 ∂u 
 ∂u  
− σ ij  i  − σ ij , j  i   dA

 ∂ε ij ∂x1
 ∂x1 , j
 ∂x1  
A
I=
leading to
A
(6.5)
We simplify Eq. (6.5) by realizing that for an elastic material (linear or non-linear)
s ij =
∂W
∂e ij
(6.6)
and the equilibrium equations gives s ij , j = 0 . The equation is simplified to:
I=
∂e ij
Ê
Ú ÁË s ij ∂x1 - s ij
A
∂ui , j ˆ
dA
∂x1 ˜¯
(6.7)
For further simplification, we note that:
s ij
∂e ij
∂x1
=
(
∂ ui , j + uj , i
1
s ij
2
∂x1
)
(6.8)
and owing to the symmetry of s ij
s ij
∂u j , i
∂x1
= s ji
∂uj , i
∂x1
= s ij
∂ui , j
∂x1
Then, Eq. (6.8) is simplified to:
s ij
∂ e ij
∂ x1
= s ij
∂ui , j
∂x1
Substituting in Eq. (6.7), we have:
I=
Ú
A
∂ui , j ˘
È ∂ui , j
- s ij
Ís ij
˙ dA Æ 0
∂x1
∂x1 ˚
Î
This proves that the left hand side of Eq. (6.4) is zero for a closed curve and is valid for all materials
whose constitutive equation can be defined through the relation of Eq. (6.6). This result will now be
124
Elements of Fracture Mechanics
used to prove the path independence of the J-Integral. Consider an area A, shown in Fig. 6.5, which is
closed by four segments, BCD, DE, EFG and GB. Note that the segments DE and GB coincide with the
crack surfaces. Segments BCD and EFG can be chosen arbitrarily within the material as long as they do
not cross each other. We now split the integral of Eq. (6.4) into four parts, as:
Ú + Ú + Ú +Ú
BCD
DE
EFG
=0
GB
A
F
D
B
E
G
C
Crack
Fig. 6.5 Area A enclosed by two curves around the crack tip and two segments
along the crack surfaces
Integrations along DE and GB do not contribute because dx2 ; 0 and Ti = 0 for the traction free crack
surfaces considered here. Two surviving integrals are:
Ú
BCD
+
Ú
=0
(6.9)
EFG
The integral along BCD is counter-clockwise with its normal pointing into the external region
and the integral along EFG is clockwise, with its normal pointing towards the interior. We can
reverse the direction of the path of the inside curve from clockwise to counter-clockwise direction
by introducing a negative sign. We then have:
Ú
=
Alternatively,
∫
=
Γ1
Ú
GFE
BCD
∫
Γ2
as shown in Fig. 6.6. Thus, we conclude that path G of the J-Integral can be chosen anywhere within
the material.
We should note here that path independence of the J-Integral is valid only for cases in
which s ij = ∂W / ∂e ij is applicable. In fact, this is a constitutive relation and thus the property of
the material plays an important role. This constitutive relation is valid for an elastic material, i.e.,
as the load is decreased, the strain retraces the path of loading (it goes from B to A as shown in
Fig. 6.1). However, this is not true for an elastic-plastic material because unloading follows a
different path from B to D. In fact, the relation between stress and strain for elastic materials is
J-Integral
Crack
125
G2
G1
Fig. 6.6 Path independence of J-integral
unique, but it is not so for an elastic-plastic material. However, if the loads on a component are
increased monotonically (with no unloading), the mathematical behavior of an elastic-plastic
material is the same as that of any non-linear elastic material. Thus, even if a component is
deformed plastically under a monotonic loading, the equations of non-linear elasticity can be used.
Under these conditions, the J-Integral is taken as path independent for plastic deformation also.
The elastic-plastic fracture mechanics is based on this idea.
The integration path G can be chosen far away from the crack tip to avoid the plastic zone or it
may be taken very close to the crack tip so that it passes through the plastic zone. In any case, the
J-Integral remains the same, provided there is no unloading. This suggests that the J-Integral is
fully capable of characterizing a crack. Depending upon the conditions of a given problem, one can
choose the path to minimize complexities in the evaluation of the J-Integral.
For linear elastic materials the J-Integral, represents the potentiality for release of energy from
the system per unit area extension of the crack growth, and is the same as G. The proof showing
that they are same [6.3] is given in Appendix 6A, presented at the end of this chapter.
6.4 STRESS-STRAIN R ELATION
We shall now replace the elastic-plastic behavior of a material by a non-linear elastic behaviour for
cases where the load on a component increases monotonically. It would be convenient if we have
only one smooth curve, such that it behaves essentially as a linear elastic material within the elastic
limit and elastic-plastic at high stresses. Note that stress-strain relations of all material are empirical.
We should choose an appropriate form of a constitutive relation.
One of the most convenient constitutive relations has been formulated by Ramberg and Osgood.
It is appropriate to describe the behavior of most engineering materials. However, there are some
materials whose behaviour cannot be modeled by the Ramberg and Osgood relation. The relation
is used in several different forms, but a convenient equation describing Ramberg–Osgood
materials [6.4] for the one-dimensional case is:
s sn
(6.10)
+
E F
where E is the Young’s Modulus of the material. Also, n and F are the other two material constants
which are determined empirically by fitting the Ramberg–Osgood equation to an experimental
stress-strain curve. In fact, n is known as the hardening exponent and its value is unity for linear
e=
126
Elements of Fracture Mechanics
elastic materials. It is greater than one for elastic-plastic materials (monotonically loaded). An
elastic-perfectly plastic material is represented by n = •. For most of the commonly used metals,
the value of n is in the range of 5–15.
The material constant F is much larger than E, and therefore the second term in the Ramberg–
Osgood relation is negligible for small values of s and then Eq. (6.10) behaves as a linear relation.
As s increases, the second term starts dominating, but the first term still contributes to the elastic
strain. In fact, Eq. (6.10) can also be written as (Fig. 6.7):
e = ee + ep
where
s
sn
and e p =
E
F
For large values of s, the first term in Eq. (6.10) becomes small and can be neglected.
ee =
s
ep
ee
e
Fig. 6.7 Elastic strain ee and plastic strain ep
We should note that n which is an exponent is a dimensionless number, but F has the dimension
of stress to the power n. Some countries still do not follow SI unit system. In fact, an advanced
country where extensive experimental data are generated still follows FPS unit system. We should
modify Eq. (6.10) from one system of units to another.
Example 6.2 An alloy steel is represented by the equation
e=
s
s 6.8
+
30, 000 2 ¥ 10 12
where s is given in ksi. Modify the equation for s given in MPa.
Solution: In the given Ramberg–Osgood equation, the units of E and F are ksi and ksi6.8
respectively. Using the equivalence relation of 1 ksi = 6.895 MPa , we modify E and F to:
E = 30,000 ¥ (6.895)1 = 206.8 ¥ 103 MPa
F = 2 ¥ 1012 ¥ (6.895)6.8 = 1.0 ¥ 1018 (MPa)6.8 .
Thus, the Ramberg–Osgood equation in SI units becomes:
e=
where s is in MPa.
(s )6.8
s
+
206.8 ¥10 3 1.0 ¥ 1018
J-Integral
127
The material constants n and F are determined from an experimental stress-strain curve by first
converting it into a relation between stress and plastic strain and then taking its logarithm. A
straight line is fitted to the data, whose slope becomes n. If a straight line cannot be fitted to the
experimental data, the material is not represented by a Ramberg–Osgood relation.
6.5 FURTHER D ISCUSSION
ON
J-INTEGRAL
6.5.1 From a Designer’s Point of View
In the previous three sections, we have introduced the definition of the J-Integral and have
discussed about some of its useful features, such as its path independence and its physical
interpretation as an energy release rate for elastic materials. We also hinted that with a clever
methodology, the J-Integral can be applied to elastic-plastic materials keeping the complexity of
the problem under control; so much so, that it can be used by a practising engineer or a designer.
How is this achieved?
Once a crack becomes critical and starts growing, most designers and engineers are not too much
concerned about its further growth. To them, as soon as a crack starts propagating, the material is
assumed to have failed. This is analogous to designers’ attitude towards yielding. It is well known
that most of the engineering materials (e.g., mild steel) deform considerably beyond yield stress
before the final rupture takes place. Still a designer regards a component as failed as soon as any
yielding initiates. Similarly, a designer looks at any kind of crack growth with panic feelings and he
would try his best to avoid it. Thus, the designer is interested in predicting loads which are just about
to propagate a crack; i.e., if the loads are increased by small amounts, the crack is likely to grow.
The fact remains that a designer is only interested in knowing the condition of “onset of the crack
growth”. This condition is exploited when the J-Integral is applied to elastic-plastic materials. If we
do not allow the growth of a crack, then we need not account for unloading. This simplifies the
analysis considerably; elastic-plastic behavior is no longer different from the non-linear elastic
behavior, as far as mathematical equations are concerned.
This idea sounds logical and rigorous, but there is a snag or a catch. Until we allow a crack to
grow by a small distance, we would never know whether we are close to the condition at which the
crack is just about to grow. In fact, while performing experiments to determine the critical value of
the J-Integral, we do allow a small growth of the crack. And as soon as we allow the growth of the
crack, the unloading is initiated and then the J-Integral approach is not rigorous anymore. Since
better techniques are not available we compromise at this point. Initially, in the 1970s many
investigators were very skeptical about the approach of the J-Integral. But based on further
investigations and many experiments, we have realized that the compromise is acceptable.
6.5.2 Experiments to Determine the Critical J-Integral
Experiments to determine the critical J are designed in such a manner that a large amount of plastic
deformation is allowed to occur near the crack tip. In fact, the plastic zone size may be as large as
the crack length. This is in sharp contrast to experiments designed to find KIc or GIc, where the
plastic zone size is controlled to remain small in comparison to the crack length.
Details of the experiments are presented in Chapter 8. However, it would be useful to discuss
some salient features of the commonly used 3-point bend specimen, as shown in Fig. 6.8. The
128
Elements of Fracture Mechanics
uncracked ligament b is chosen to be approximately the same as the crack length a. The bending
moment is maximum at the centre of the span, where the depth of the beam is considerably smaller
than W. Consequently, stresses are very high in front of the crack tip, forming a plastic hinge and
the rotation of the beam, as shown in the figure. It is to be noted here that in the rest of the beam,
elastic deformation is small, owing to combined effect of larger depth and lower bending moment.
Consequently, the deflection u is caused mainly by the plastic deformation at the crack tip. Thus,
clever experimental techniques have been developed which account for the large plastic zone in
the vicinity of the crack tip.
Fig. 6.8 Plastic hinge formed in the uncracked ligament b
6.5.3 Comments on the Numerical Evaluation of J-Integral
As stated earlier, the primary goal of a designer is to predict whether a crack in a component is likely
to grow under the given loading conditions. He can either obtain the critical J value from literatures
or he can conduct an experiment to determine it. Stress-strain relations, such as the Ramberg–Osgood
relation describing the elastic-plastic material (only for monotonically increasing load) are also
available to the designer. His task is to evaluate the J-Integral in an application knowing the
constitutive relations, the loading conditions and, of course, the geometry of the component.
The J-Integral is generally evaluated through a numerical programme, most probably using a
finite element method. With the availability of sophisticated computers, the task is not difficult.
The designer may take the help of an expert in the field of FEM. These days standard packages are
also available, which makes it simpler for a designer to determine the value of J. This is because
the J-Integral is evaluated by integrating on an arbitrary path around the crack tip and we need not
evaluate the strain or stress field very accurately near the crack tip. In fact, we can choose the
integration path which is far away from the crack, thus avoiding the use of singular elements in the
vicinity of the crack tip. Also, we do not require choosing a fine grid close to the crack tip and still
maintain the calculations reasonably accurate. With the increasing acceptance of finite element
methods and the availability of standard packages, the determination of J is no longer a
prohibitively difficult task in research and development work of many industries.
6.5.4 Predicting Safety or Failure
The J-Integral may be determined through a numerical method, or it may also be obtained from a
handbook. In an application, if the J value of a crack is less than the critical J-Integral (Jc), the design
J-Integral
129
is safe against any fracture failure. On the other hand, whenever J exceeds Jc the designer must modify
his design, either by choosing a tougher material or changing the geometry of the component. For
example, if the failure is predicted for a slot in a component, the length of the slot may be reduced. In
other cases, the stress on the component may be decreased by changing the geometry, such as
increasing the moment of inertia of the component. In some cases, the designer may not have the
freedom to change the geometry and he may settle for lower limits on the external loads.
6.5.5 Comments on the Experimental Determination of the
Toughness of Ductile Materials
The usage of the SIF method is mostly restricted to the LEFM, where the plastic zone is considerably
smaller than the crack length. Experiments to determine the critical SIF are carefully designed to
restrict the size of the plastic zone. This is done by making the specimen thick, so as to achieve plane
strain conditions. Also, the lateral dimensions of the specimen are chosen reasonably large, so that a
large portion of the specimen is under elastic deformation. Because the plastic zone of some tough
materials is fairly large, the specimen size becomes accordingly larger, so much so that in certain
cases the specimen size should be made of a plate as thick as 200 mm or more. A specimen of such
thickness may also create many problems. First of all, the loads required to deform such a large
specimen are enormously high and the test would require a machine of an extremely high capacity
and also a high cost. Secondly, large amounts of the material will be required, which may have to be
made specially of such thickness. Thirdly, the machining and handling cost of specimen will become
rather steep. Because of these practical problems, Begley and Landes looked for alternative
techniques in determining material toughness and suggested that the J-Integral provides more
feasible and economical test-techniques for determining the toughness of ductile materials [6.5].
The J-Integral approach, on the other hand, allows the analysis of any elastic-plastic behavior of
the material in the vicinity of the crack tip. Unlike the SIF methods, this approach does not limit the
size of the plastic zone during the analysis, as the non-linear material behavior is incorporated. To
determine the critical J-Integral, the specimens are designed in such a manner that there results a
large amount of plastic deformation near the crack tip, so much so that the elastic deformation in
the rest of the specimen may be neglected in many experiments. There is no need to make the
specimen very large and the critical J value can be determined in a laboratory equipped with a
normal loading machine, such as a 10 ton tensile machine.
In spite of several strong points in favor of the J-Integral approach, the SIF method is still
surviving. Unlike the critical K whose values are listed in literature for commonly used engineering
materials, the critical J is still not widely available in literature and handbooks. If a designer is
adopting the J-Integral approach, he may have to pay for the experimental tests, which usually
turn out to be rather expensive.
Moreover, it has been found that there is a large variation in the critical J value for ductile
materials, due to its non-linear stress-strain behavior. For a small variation in the critical load, the
corresponding variation in Jc is much larger. However, the large variation in Jc should not disturb
us, because whenever a designer predicts loads for a maximum allowable J, the variation in the
failure-load will be accordingly smaller. For example, two tests on the same material may give the
critical J value with a large variation of 40% or so. When these results are used by a designer,
maximum allowable load may vary only by 5%. There is nothing wrong in such an analysis,
provided we realize that the large variation in the value of the critical J is normal.
130
Elements of Fracture Mechanics
6.6 ENGINEER A PPROACH—A SHORT C UT
6.6.1 A Simplified Relation for the J-Integral
To minimize errors designers like to avoid making lengthy calculations and complex analysis. They
would rather make use of the formulae available in handbook and plug numbers to get the answer.
Therefore, natural question arises—can we prepare tables or handbooks that would avoid
numerical analysis to determine J value of a crack?
The energy release rate in a linear elastic case can be expressed as:
b 2 K 2 b 2 ps 2 a
(6.11)
=
E
E
where b accounts for the geometry of the crack and the component. For example, b 2 = 1 in a centrecracked infinite plate, but for an edge crack, b 2 becomes 1.12. For other cases, b 2 may be taken
G=
from a handbook. Substituting e = s / E in Eq. (6.11), we obtain:
G = b 2p s e a
(6.12)
For materials following the Ramberg–Osgood relation, it has been realized that an expression [6.4]
analogous to elastic cases [Eq. (6.12)] can be developed as:
Jp = H s ep a
(6.13)
where ep is the plastic strain and H is a dimensionless factor that depends on geometry. Taking the
plastic strain from Eq. (6.10) as ep = sn/F and substituting the same in Eq. (6.13), we obtain:
H s n+1a
F
Then overall the J-Integral becomes:
Jp =
J=
(6.14)
p b 2 s 2 a H s n + 1a
+
F
E
(6.15)
If handbooks/tables listing b2 and H values are made available to a designer and material
constants E, n and F are known, the designer can easily determine the values of J, thus
minimizing the chances of making errors in computing J through a numerical analysis. In case,
the value of s is large, the second term in Eq. (6.15) dominates and we may ignore the effect of the
elastic deformation.
Example 6.3 Determine the J-Integral for a component loaded in Mode I, with a far field stress of
200 MPa and an edge crack of 40 mm length. The geometrical factors are b 2 = 1.12 and H = 7; the
material follows the Ramberg–Osgood relation with the material constants given as:
E = 207 GPa, n = 6.8, and F = 1 ¥ 1018 (MPa)6.8
Solution: Substituting the data in Eq. (6.15), we obtain:
J=
p ¥ 1.12 ¥ (200 ¥ 106 Pa)2 ¥ (0.04 m)
(207 ¥ 109 Pa)
= 27.2 + 248.4 = 275.6 kJ/m2
+
7 ¥ [(2007.8 (MPa)7.8 ] ¥ (0.04 m)
[1 ¥ 1018 (MPa)6.8 ]
J-Integral
131
Knowing the critical J-Integral (Jc) through experiments and using Eq. (6.15), a designer may like
to determine the maximum allowable stress s. The equation can be solved for s through an
iterative procedure by guessing its value in the first trial. The problem can be solved easily with the
help of a digital computer. In case the elastic contribution is negligible, s is directly obtained as:
È FJ ˘
s max = Í c ˙
Î Ha ˚
1/( n + 1)
(6.16)
To study how Ds max varies with D Jc, we differentiate the equation and then divide it by s max to have:
Ds max Ê 1 ˆ D J c
=Á
Ë n + 1˜¯ J c
s max
We note that usually n is significantly larger than one and therefore the percentage variation
in smax is much less than the percentage variation in Jc.The above relation justifies the discussion
presented in Sec. 6.5.6. We may look at the problem from a different angle. If Jc is to be determined
experimentally, smax may easily vary by 5–10% from specimen to specimen. The variation in Jc
will be about eight times larger if n is close to 7. Therefore, we conclude that large variation in
the experimental values of Jc should not be seen with contempt; the variation in the predicted
smax values would still be small.
6.6.2 Applications to Engineering Problems
For any application of the J-Integral to practical problems, the Ramberg–Osgood relation is written
in a slightly different form [6.3] as:
Ès ˘
e
s
+a Í ˙
=
s0
e0
Îs o ˚
n
(6.17)
This relation has four material constants (n, a, s0 and e0) in place of the three constants of Eq. (6.10).
However, this form of the relation is better, because a is a dimensionless number. Since an equation
expressed in another form cannot have more independent constants, there should be some relation
between the four constants. In fact, s0 can be chosen anywhere on the elastic portion of the s – e
curve and the corresponding e0 is evaluated by the relation:
s0 = Ee0
(6.18)
Many investigators prefer to choose s0 as the yield stress (sys) of the material. Realizing a relation
exists between s0 and e0 , we can compare Eq. (6.17) to Eq. (6.10) and obtain F =s 0n / e 0a .
Corresponding to Eq. (6.17), Jp is usually expressed as [6.6]:
Ê Pˆ
Jp = as 0 e 0 bg1h1 Á ˜
Ë P0 ¯
n +1
(6.19)
where P is the applied load per unit thickness of plate, P0 is the limit or collapsed load of the plate
based on yield stress sys (=s0), b is the uncracked ligament; and g1 and h1 are geometric factors
which depend on a/W and n. Values of g1 and h1 are listed for some common practical applications
in Appendix 6B, presented at the end of this chapter [6.3, 6.7].
132
Elements of Fracture Mechanics
For a centre-cracked plate shown in Fig. 6.9, P0 depends on the collapse-load for the material
yielding in the uncracked ligament b on each side, and is given by:
P0 = 2 bs 0
P0 =
for plane stress
4
bs 0
3
for plane strain.
s
2a
b
b
s
2W
Fig. 6.9 Centre-cracked plate
Example 6.4 A centre-cracked panel of width 2W = 500 mm and thickness B = 20 mm is pulled
normal to the crack length (2a = 100 mm), with a far field stress s. Estimate the maximum s that can
be applied without causing the growth of the crack if, Jp = 400 kJ/m2. The material constants of the
Ramberg–Osgood equation are known to be n = 5, a = 5.4, sys (= s0) = 520 MPa and e0 = 0.00251.
Solution: We would use the equation:
Ê Pˆ
Jp = a s 0 e 0 b g1 h1 Á ˜
ËP ¯
n +1
0
(500 mm - 100 mm)
= 200 mm
2
To find P0, g1 and h1 through the tables of Appendix 6B, we should know whether the plate is
subjected to plane strain or plane stress. At this stage, we start solving the problem, assuming that
the plane strain conditions prevail. Once the maximum s is determined, we can evaluate the plastic
zone size and check the justification of our assumption.
From Appendix 6B, we obtain:
where,
b=
P0 =
4bs 0 4 ¥ (0.2 m) ¥ (520 ¥ 106 Pa)
=
= 240 ¥ 106 N/m
3
3
J-Integral
g1 =
133
50 mm
a
=
= 0.2
W 250 mm
h1 = 3.71
Substituting these values, we obtain:
Jp
È
˘
P
= Í
˙
P0
Î a s 0 e 0 bg1 h1 ˚
1/( n + 1)
1/6
È
˘
(400 ¥ 103 J/m2 )
= Í
˙
6
Î 5.4 ¥ (520 ¥ 10 Pa) ¥ 0.00251 ¥ (0.2 m) ¥ 0.2 ¥ 3.71 ˚
= 0.852
which gives,
P = 0.852 ¥ (240 ¥ 106 N/m) = 204.5 ¥ 106 N/m
s=
204 ¥ 106 N/m
P
=
= 408 MPa
2W
(0.5 m)
We should now explore whether our initial assumptions regarding this problem as a case of
plane strain were justified. Therefore, an estimate should be made on the plastic zone size. Earlier
in Chapter 5, the Dugdale Approach was found to be handy, but in that case the material was taken
to be elastic-perfectly plastic. With a hardening material, the plastic zone size (rp) is obviously
smaller, but the estimation of rp is difficult. However, analysis of Mode III is simpler and rp has
been determined for a work-hardening material with the hardening exponent n. Based on its
results, rp of Mode I is estimated to be:
rp =
1 (n - 1) KI2
2
lp ((n + 1) s ys
where, l = 1 for plane stress, and l = 3 for plane strain. KI is evaluated on the basis of the actual
crack length to keep the calculations simple, using the relation:
KI = s p a f (a )
where,
50 mm
a
=
= 0.2
W 250 mm
f (a) = 1.0 + 0.128a - 0.288a2 + 1.523a3 = 1.026
a=
Then, we have:
KI = (408 MPa) x p1/2 ¥ (0.05 m)1/2 ¥ 1.026
= 165.9 MPa m
The plastic zone size for a plane strain is:
2
rp ª
1 4 Ê 165.9 MPa m ˆ
¥
= 7.20 mm
3p 6 ÁË 520 MPa ˜¯
The plastic zone size is only slightly smaller than the thickness of the plate. We should now redo
the entire problem for the case of a plane stress:
134
Elements of Fracture Mechanics
P0 = 2bs ys
= 2 ¥ (0.2 m) ¥ (520 ¥ 106 Pa) = 208 ¥ 106 N/m
h1 = 3.71
P
È
˘
400 ¥ 103 J/m 2
= Í
˙
6
P0
Î 5.4 ¥ (520 ¥ 10 Pa) ¥ 0.00251 ¥ 0.2 ¥ (0.2m) ¥ 3.71 ˚
1/6
= 0.852
P = 0.852 ¥ 208 ¥ 106 = 177.2 ¥ 106 N/m
s=
P 177.2 ¥ 106
=
= 354.4 MPa
2W
0.5
which gives the SIF as:
KI = 354.4 ¥ p 1/2 ¥ (0.05)1/2 ¥ 1.026 = 144 MPa m
The plastic zone size for plane stress is estimated to be:
2
rp =
1 (n - 1) KI2 1 4 Ê 144 MPa m ˆ
= ¥ ¥Á
= 16.27 mm
2
p (n + 1) s ys
p 6 Ë 520 MPa ˜¯
This is close to the plate thickness (= 20 mm) and therefore the plane stress analysis is applicable.
Example 6.5 Consider an axially cracked pressured cylinder of steel (E = 207 GPa) with internal
radius of 80 mm and wall thickness of 16 mm (Fig. 6.10). An axial crack of 3 mm depth has been
identified on the inner surface of the cylinder. If the yield stress s 0 of the material is 700 MPa,
the hardening exponent n = 7, the material constant of Ramberg–Osgood relation a = 6.2 and Jp = 280
J/m2, determine the maximum pressure that the cylinder can resist against the crack growth.
80
16 mm
p
3 mm
Fig. 6.10 Axially cracked pressurized cylinder
Solution:
Ri = 80 mm, W = 16 mm, a = 3 mm, b = 13 mm,
Rc = 83 mm, a/W = 0.188, W/Ri = 1/5 (Appendix 6B, Fig. 6.17)
n = 7, a = 6.2
J-Integral
135
sys = s0 = 700 MPa, Jp = 280 kJ/m2
e0 = s 0 / E =
700
= 0.00338
207 ¥ 103
From Appendix 6B, we obtain:
a
g1 =
= 0.188
W
h1 = 11.4
p0 =
2bs ys
3 Rc
=
2 ¥ (0.013m ¥ 700 MPa)
3 ¥ ( 0.083 m )
= 126.6 MPa
Then, we have:
Jp
Ê
ˆ
p
= Á
˜
bg
h
p0
a
s
e
Ë
0
0
1 1¯
1/( n + 1)
Ê
ˆ
280 ¥ 103 J/m 2
= Á
6
Ë 6.2 ¥ (700 ¥ 10 Pa) ¥ 0.00338 ¥ (0.013m) ¥ 0.188 ¥ 11.4 ˜¯
1/8
= 0.953
leading to
p = 0.953 ¥ 126.6 = 120.6 MPa
6.7 CLOSURE
The plastic zone size near the tip of a crack is large for tough materials, and parameters G and K of
LEFM may not be suitable. Therefore, another parameter, the J-Integral has been developed that
can also account for non-linear stress-strain behaviour of the plastic zone. Such analysis is known
as (EPFM).
The J-Integral is an expression which remains path independent when integrated along a path
from any point of a crack surface to any point of the other crack surface. When the J-Integral is
applied to a linear elastic material, it is same as the energy release rate G.
To account for the elastic-plastic behavior, only a single constitutive relation is generally
adopted. It has been found that the Ramberg–Osgood stress-strain relation can be employed for
most engineering materials. At low stresses, the equation is almost linear, representing elastic
deformation; at high stresses, the non-linear behavior dominates.
For checking whether a load is safe on a component, the J-Integral can be evaluated either
through numerical analyses or by using tables from a handbook. If the value of J of a crack is less
than the critical Jc, the material is safe against any crack growth.
The experimental procedure to determine the value of JIc is described in Chapter 8. It is worth
mentioning that the plate thickness of a specimen for finding KIc is large and impractical for
materials with the large plastic zone, because plane strain conditions must be created. On the
contrary, a test specimen to determine JIc allows a large plastic zone at the crack tip and the
specimen thickness does not necessarily have to be large.
136
Elements of Fracture Mechanics
APPENDIX 6A
Equivalence of G and J for Elastic Materials
It would be proved that for plane bodies, made of linear elastic materials, the expression of the
J-Integral is equivalent to G. Consider the general case of a plane body of area A with no body
forces. The contour of this body is represented with a curve G0, which is split into two parts: GT,
where traction is prescribed and Gu, on which displacement conditions are given (Fig. 6.11). Note
that the free surface belongs to GT. Potential energy F is then given by:
G0 = G u + G T
GT
G0
Ti
Gu
Fig. 6.11 Path G0 chosen to coincide with the body contour and cracked surfaces
F=
Ú WdA - Ú
Ti ui ds
GT
A
Leading to,
G= -
du
dF
dW
dA + Ú Ti i ds
= -Ú
da
da
da
A
G
(6.20)
T
Transforming the origin to the crack tip by the relation,
X1 = x1 – a
we obtain:
d
∂ ∂X ∂
= + 1
da ∂a ∂a ∂X1
From Eq. (6.21), we have:
∂
∂
∂X 1
= -1; and
=
∂
X
∂
x1
∂a
1
Then, the differentiation relation simplifies to:
∂
∂
d
= ∂a - ∂x
1
da
(6.21)
J-Integral
137
Operating it on Eq. (6.20), we obtain:
G = -Ú
A
∂W
∂W
∂u
∂u
dA + Ú
dA + Ú Ti i ds - Ú Ti i ds
x
a
x1
∂a
∂
∂
∂
1
A
G
G
T
(6.22a)
T
But the integrand of the first term is modified as follows:
∂W ∂e ij 1 ∂(ui , j + uj ,i )
∂W
=
= s ij
∂e ij ∂ a 2
∂a
∂a
Making use of the symmetry of stress tensor (s ij = s ji ) , we obtain:
∂(ui , j )
Ê ∂u ˆ
∂W
= s ij Á i ˜
= s ij
Ë ∂a ¯ i , j
∂
a
∂a
(6.22b)
Invoking the principle of virtual work, we have:
Ú
GT
Ti
∂ui
ds =
∂a
È ∂u ˘
Ú s ij ÍÎ ∂ai ˙˚
A
dA
(6.22c)
i, j
Substituting Eqs (6.22b) and (6.22c) in Eq. (6.22a), we obtain (note ∂u i = 0 on G u and therefore
G T Æ G 0)
G=
∂W
Ú ∂x
A
1
dA - Ú Ti
G0
∂u
ds
∂x1
(6.23)
The first term can now be converted into the line integral by invoking the Divergence Theorem,
for which closed contour G0 is chosen to coincide with the body contour and the crack faces as
shown with the dashed line in Fig. 6.11. Although the singularity of the crack tip lies on the path G0,
it is not serious for the first term of Eq. (6.23). For small values of r, strain energy density W varies,
as 1/r because it is a product of stress and strain, each varying as 1/ r . Therefore, dW/dx1 varies
as 1/r, but dA varies a r2 and thus the singularity is nullified when r Æ 0. Then, the divergence
theorem modifies Eq. (6.23) to:
G=
ÚWn
1
G0
ds - Ú Ti
G0
∂ui
ds
∂x1
The contour G0 can be split into two parts,
G0 = G + Gc
where Gc is the portion on the crack surfaces and G is the usual integration path of the J-Integral
(from one crack surface to another). Also, it is worth noting that on Gc path dx2 ª 0 and Ti = 0. By
noting n1ds = dx 2 [Fig. 6.4(b)], the above equation is thus simplified to:
G=
∂ui
Ú W dx - Ú T ∂x
i
2
G
G
ds
1
The right hand side has the expression of J, which is path independent, and therefore,
G=J
Thus, for linear elastic materials, the expression of the J-Integral is equivalent to G.
138
Elements of Fracture Mechanics
APPENDIX 6B
The J-Integral of Some Common Cases through Engineering Approach
n +1
For the engineering approach (Sec. 6.6), Jp is defined as J p = as 0e 0 bg1 h1 ( P / P0 ) . In this appendix,
the expressions for the geometric factor g1 and collapse load P0 are given, and the geometric factor
h1 is listed for plane stress (p – s) and plane strain (p – e) for some commonly encountered cases.
Usually, s0 is chosen to be same as yield stress (sys).
6B.1 THREE-POINT B END SPECIMEN
The specimen is loaded with force P per unit thickness, as shown in Fig. 6.12.
P
b
W
a
L
L
Fig. 6.12 Three-point bend specimen
P0 = 0.728 sys b2/L for plane strain
P0 = 0.536 sys b2/L for plane stress
g1 = 1, and h1 = listed in Table 6.1 for L/W = 2.
TABLE 6.1 h1 for three-point bend specimen
a/W
Type
1/8
p-e
p -s
p-e
p -s
p-e
p -s
p-e
p -s
p-e
p -s
p-e
p -s
1/4
3/8
1/2
5/8
3/4
n
1
2
3
5
7
10
13
16
20
0.937
0.676
1.20
0.869
1.33
0.963
1.41
1 .02
1.46
1.05
1.48
1.07
0.869
0.600
1.034
0.731
1.15
0.797
1.09
0.767
1.07
0.786
1.15
0.786
0.805
0.548
0.930
0.629
1.02
0.680
0.922
0.621
0.896
0.649
0.974
0.643
0.687
0.459
0.762
0.479
0.846
0.527
0.675
0.453
0.631
0.494
0.693
0.474
0.580
0.383
0.633
0.370
0.695
0.418
0.495
0.324
0.436
0.357
0.500
0.343
0.437
0.297
0.523
0.246
0.556
0.307
0.331
0.202
0.255
0.235
0.348
0.230
0.329
0.238
0.396
0.174
0.442
0.232
0.211
0.128
0.142
0.173
0.223
0.167
0.245
0.192
0.304
0.117
0.360
0.174
0.135
0.0813
0.084
0.105
0.140
0.110
0.165
0.148
0.215
0.0593
0.265
0.105
0.0741
0.0298
0.411
0.471
0.0745
0.0442
J-Integral
6B.2 CENTRE-CRACKED P LATE
The plate is pulled by the load P per unit thickness, as shown in Fig. 6.13.
P
b
a
W
P
Fig. 6.13 A centre-cracked plate
P0 = 4b s ys / 3
for plane strain
P = 2b s ys
for plane stress
g1 = a/W
h1 = (Listed in Table 6.2)
TABLE 6.2 h1 for centre cracked plate
n
a/W
Type
1
2
3
5
7
10
13
16
20
1/8
p-e
p -s
2.80
2.80
3.61
3.58
4.06
4.01
4.35
4.47
4.33
4.65
4.02
4.62
3.56
4.41
3.06
4.13
2.46
3.72
1/4
p-e
p -s
2.54
2.54
3.01
2.97
3.21
3.14
3.29
3.20
3.18
3.11
2.92
2.86
2.63
2.65
2.34
2.47
2.03
2.20
3/8
p-e
p -s
2.34
2.34
2.62
2.53
2.65
2.52
2.51
2.35
2.28
2.17
1.97
1.95
1.71
1.77
1.46
1.61
1.19
1.43
1/2
p-e
p -s
2.21
2.21
2.29
2.20
2.20
2.06
1.97
1.81
1.76
1.63
1.52
1.43
1.32
1.30
1.16
1.17
0.978
1.00
5/8
p-e
p -s
2.12
2.12
1.96
1.91
1.76
1.69
1.43
1.41
1.17
1.22
0.863
1.01
0.628
0.853
0.458
0.712
0.300
0.573
3/4
p-e
p -s
2.07
2.07
1.73
1.71
1.47
1.46
1.11
1.21
0.895
1.08
0.642
0.867
0.461
0.745
0.337
0.647
0.216
0.532
139
140
Elements of Fracture Mechanics
6B.3 COMPACT TENSION SPECIMEN
The standard compact tension specimen is pulled by the force P, as shown in Fig. 6.14.
P
+
+
P
b
a
W
Fig. 6.14 Compact tension specimen
P0 = 1.455 b b s ys
for plane strain
P0 = 1.071 b b s ys
for plane stress
g1 = 1
h1 = (Listed in Table 6.3)
b = [(2a/b)2 + 4a/b + 2]1/2 – 2a/b – 1 .
TABLE 6.3 h1 for compact tension specimen
a/W
Type
1/4
n
1
2
3
5
7
10
13
16
20
p-e
p -s
2.23
1.61
2.05
1.47
1.78
1.28
1.48
1.06
1.33
0.903
1.26
0.729
1.25
0.601
1.32
0.511
1.57
0.395
3/8
p-e
p -s
2.15
1.55
1.72
1.25
1.39
1.05
0.970
0.801
0.693
0.647
0.443
0.484
0.276
0.377
0.176
0.284
0.098
0.220
1/2
p-e
p -s
1.94
1.40
1.51
1.08
1.24
0.901
0.919
0.686
0.685
0.558
0.461
0.436
0.314
0.356
0.216
0.298
0.132
0.239
5/8
p-e
p -s
1.76
1 .27
1.45
1.03
1.24
0.875
0.974
0.695
0.752
0.593
0.602
0.494
0.459
0.423
0.347
0.370
0.248
0.310
3/4
p-e
p -s
1.71
1.23
1.42
0.977
1.26
0.833
1.03
0.683
0.864
0.598
0.717
0.506
0.575
0.431
0.448
0.373
0.345
0.314
J-Integral
6B.4 SINGLE-EDGE N OTCHED P LATE
The plate with an edge crack is pulled by far field stress s (= P/W) as shown in Fig. 6.15.
s
a
b
W
s
Fig. 6.15 Single-edge-notched plate
P0 = 1.455 b b s ys
for plane strain
P0 = 1.072 b b s ys
for plane stress
g1 = a/W
h1 = (Listed in Table 6.4)
b = [ 1+(a/b)2]1/2 – a/b
TABLE 6.4 h1 for single-edge-notched plate
a/W
Type
1/8
p-e
p -s
p-e
p -s
p-e
p -s
p-e
p -s
p-e
p -s
p-e
p -s
1/4
3/8
1/2
5/8
3/4
n
1
2
3
5
7
10
13
16
20
4.95
3.58
4.34
3.14
3.88
2.81
3.40
2.46
2.86
2.07
2.34
1.70
6.93
4.55
4.77
3.26
3.25
2.37
2.30
1.67
1.80
1.41
1.61
1.14
8.57
5.06
4.64
2.92
2.63
1.94
1.69
1.25
1.30
1.106
1.25
0.910
11.5
5.30
3.82
2.12
1.68
1.37
0.928
0.776
0.697
0.755
0.769
0.624
13.5
4.96
3.06
1.53
1.08
1.01
0.514
0.510
0.378
0.551
0.477
0.447
16.1
4.14
2.17
0.960
0.539
0.677
0.213
0.286
0.153
0.363
0.233
0.280
18.1
3.29
1.55
0.615
0.276
0.474
0.090
0.164
0.0625
0.248
0.116
0.181
19.9
2.60
1.11
0.400
0.142
0.342
0.0385
0.0956
0.0256
0.172
0.059
0.118
21.2
1.92
0.712
0.230
0.0595
0.226
0.0119
0.0469
0.0078
0.107
0.0215
0.0670
141
142
Elements of Fracture Mechanics
6B.5 DOUBLE-E DGE N OTCHED PLATE
The double-edge notched plate is pulled by far field stress s (= P/2W), as shown in Fig. 6.16.
s
b
a
b
a
2W
s
Fig. 6.16 Double-edge-notched plate
P0 = (0.72W + 1.82b) s ys
for plane strain
P0 = 4b s ys/ 3
g1 = 1
h1 = (Listed in Table 6.5)
for plane stress
TABLE 6.5 h1 for double-edge-notched plate
a/W
Type
1/8
n
1
2
3
5
7
10
13
16
20
p-e
p -s
0.572
0.583
0.772
0.825
0.922
1.02
1.13
1.37
1.35
1.71
1.61
2.24
1.86
2.84
2.08
3.54
2.44
4.62
1/4
p-e
p -s
1.10
1.01
1.32
1.23
1.38
1.36
1.65
1.48
1.75
1.54
1.82
1.58
1.86
1.59
1.89
1.59
1.92
1.59
3/8
p-e
p -s
1.61
1.29
1.83
1.42
1.92
1.43
1.92
1.34
1.84
1.24
1.68
1.09
1.49
0.970
1.32
0.873
1.12
0.674
1/2
p-e
p -s
2.22
1.48
2.43
1.47
2.48
1.38
2.43
1.17
2.32
1.01
2.12
0.845
1.91
0.732
1.60
0.625
1.51
0.208
5/8
p-e
p -s
3.16
1.59
3.38
1.45
3.45
1.29
3.42
1.04
3.28
0.882
3.00
0.737
2.54
0.649
2.36
0.466
2.27
0.020
3/4
p-e
p -s
5.24
1.65
6.29
1.43
7.17
1.22
8.44
0.979
9.46
0.834
10.9
0.701
11.9
0.630
11.3
0.297
17.4
J-Integral
143
6B.6 AXIALLY CRACKED P RESSURIZED CYLINDER
Pressurized cylinders are commonly used in industries; a crack along the axial direction may grow
under pressure po, as shown in Fig. 6.17.
i
R
W
p
a
b
Fig. 6.17 Axial crack in a pressurized cylinder
p0 = 2b s ys /( 3 Rc)
g1 = a/W
h1 = (Listed in Table 6.6)
Rc = (Ri + a) .
TABLE 6.6 h1 for internally pressurized cylinder with axial crack
a/W
W/Ri
1/8
n
1
2
3
5
7
10
1/5
1/10
1/20
6.32
5.22
4.50
7.93
6.64
5.79
9.32
7.59
6.62
11.5
8.76
7.65
13.1
9.34
8.07
14.9
9.55
7.75
1/4
1/5
1/10
1/20
7.00
6.16
5.57
8.34
7.49
6.91
9.03
7.96
7.37
9.59
8.08
7.47
9.71
7.78
7.21
9.45
6.98
6.53
1/2
1/5
1/10
1/20
9.79
10.5
10.8
10.4
11.6
12.8
9.07
10.7
12.8
5.61
6.47
8.16
3.52
3.95
4.88
2.11
2.27
2.62
3/4
1/5
1/10
1/20
11.00
16.1
23.1
5.54
8.19
13.1
2.84
3.87
5.87
1.24
1.46
1.90
0.83
1.05
1.23
0.493
0.787
0.883
144
Elements of Fracture Mechanics
6B.7 CIRCUMFERENTIALLY C RACKED C YLINDER
Figure 6.18 shows the cylinder subjected to an axial load P, with a circumferential crack on the
inside surface of depth a. The far field axial stress for this case is s = P /( p ( R02 - Ri2 )).
s
R0
Ri
b
a
W
s
Fig. 6.18 Circumferentially cracked cylinder
P0 = 2 p s ys (R02 - Rc2 )/ 3
g1 = a/W
h1 = (Listed in Table 6.7)
Rc = Ri + a
TABLE 6.7 Value of h1 for circumferentially cracked cylinder
a/W
W/Ri
1/8
n
1
2
3
5
7
10
1/5
1/10
1/20
3.78
4.00
4.04
5.00
5.13
5.23
5.94
6.09
6.62
7.54
7.69
7.82
8.99
9.09
9.19
11.1
11.1
11.1
1/4
1/5
1/10
1/20
3.88
4.17
4.38
4.95
5.35
5.68
5.64
6.09
6.45
6.49
6.93
7.29
6.94
7.30
7.62
7.22
7.41
7.65
1/2
1/5
1/10
1/20
4.40
5.40
6.55
4.78
5.90
7.17
4.59
5.63
6.89
3.79
4.51
5.46
3.07
3.49
4.13
2.34
2.47
2.77
3/4
1/5
1/10
1/20
4.12
5.18
6.64
3.03
3.78
4.87
2.23
2.57
3.08
1.55
1.59
1.68
1.30
1.31
1.30
1.11
1.10
1.07
J-Integral
145
QUESTIONS
1. Why is the numerical evaluation of JI usually simpler than the evaluation of GI or KI, in the
case of the LEFM?
2. Path independence of the J-Integral is not valid for elastic-plastic materials. Why?
3. If path independence is not valid for elastic-plastic materials, how can we apply it to real
life cases dealing with large plastic zone in the vicinity of the crack tip?
4. Although a variation is found to be large in an experimentally determined JIc , it is
acceptable for design purposes. Justify.
5. Why is the Ramberg-Osgood relation convenient for determining the J-Integral for elasticplastic materials?
6. Why are the specimens very thick to determine the value of KIc of ductile materials?
7. Unlike the case of KIc-test, the specimen recommended for finding the critical J-Integral
need not be thick for ductile materials. Why?
PROBLEMS
1. Consider an infinitely long strip with an edge-crack of length a. The lower surface of the
strip is bonded to a rigid surface and the upper surface to a rigid rail as shown in Fig. 6.19.
Rigid rail
x2
u
E
F
A
B
x1
D
h
C
h
a
Fig. 6.19 The figure of Problem 1
If the rail is displaced in the x2 direction by distance u, determine JI (and then KI) by choosing
a convenient path. Consider plane stress and plane strain cases separately. Note that:
(a) Path ABCDEF is so chosen that points C and D are quite far away from the crack tip,
(b) The variation of the displacement field with x1 is zero, and
(c) On segment CD,
Plane stress: s11 π 0, s 22 π 0, s 33 = 0, e11 = 0, e22 π 0, e33 π 0;
Plane strain: e11= 0, e22 π 0, e33 = 0, s 11 π 0, s 22 π 0, s 33 π 0
2. Consider two infinitely long strips of thickness h1 and h2 with material properties as shown
in Fig. 6.20. These strips are bonded together with an edge-crack of length a. The strips are
wide enough to assume plane strain conditions. The lower face of the bottom strip is
bonded to a rigid surface, while the top surface of the upper strip is bonded to a rigid rail.
Determine the J-Integral, if the rail is pulled up by distance u. (Note: At the interface
(1)
(2)
(1)
(2)
= s 22
π e 22
(s 22
but e 22
.)
Elements of Fracture Mechanics
x2
Rigid rail
u
x1
E1, n1
h1
E2, n2
h2
Bond-interface
a
Fig. 6.20 The figure of Problem 2
3. Same as Problem 1, with a difference that the top rail is moved by a distance w in the x3
direction.
4. Consider a DCB specimen made by bonding two infinitely long strips as shown in Fig. 6.21.
Determine JI if the ends of cantilevers are loaded with moments M.
x2
M
E1, n1
x1
M
E2, n2
h1
h2
Interface
a
Fig. 6.21 The figure of Problem 4
5. Stress-strain relation for a steel alloy with modulus E = 207 GPa is shown in Fig. 6.22.
Determine sys, and material constants a and n of the Ramberg–Osgood equation.
2000
1600
1200
s, MPa
146
800
400
0
0.00
0.20
0.40
0.60
e
Fig. 6.22 The figure of Problem 5
0.80
J-Integral
147
6. Consider a three-point bend specimen with a centre load as shown in Fig. 6.23. The material
properties for the Ramberg-Osgood relation are:
Fig. 6.23 The figure of Problem 6
sys = s0 = 700 MPa, e0 = s0/E
E = 207 GPa, a = 8.2, n = 6
(a) Determine KI
(b) Estimate the plastic zone size
(c) Determine GI based on the LEFM
(d) Determine Jp using the engineering approach.
7. Consider a circumferentially cracked cylinder of 240 mm internal radius, 24 mm wall
thickness and 3 mm long crack all around the circumference (Fig. 6.18). The material follows
the Ramberg-Osgood equation with the following material constants:
sys = s0 = 800 MPa, e0 = s0/E
E = 207 GPa, a = 8.6, n = 3.
The cylinder is subjected to axial tension. If Jp = 150 kJ/m2, determine the maximum axial
tensile stress. Neglect Je.
8. Determine the maximum tensile stress for the cylinder of Problem No. 7 when the wall
thickness and crack length are changed to 8 mm and 2 mm respectively. All other
dimensions and material properties remain the same.
REFERENCES
6.1 Chakrabarty, J. (1987). Theory of Plasticity, McGraw-Hill Book Company, New York.
6.2 Rice, J.R. (1968). A Path Independent Integral and Approximate Analysis of Strain
Concentrations by Notches and Cracks, Journal of Applied Mechanics, 35, pp. 379–386.
6.3 Kanninen M.F. and Popelar, C.H. (1985), Advanced Fracture Mechanics, Oxford University
Press, New York.
6.4 Broek, David (1988). The Practical Use of Fracture Mechanics, Kluwer Academic Publishers,
Dordrecht.
148
Elements of Fracture Mechanics
6.5 Begley, J.A. and Landes, J.D. (1972). The J-Integral as a Fracture Criterion, Fracture
Toughness, p1. Part II, ASTM STP 514, American Society for Testing and Materials,
Philadelphia, pp. 1–20.
6.6 Goldman, N.L. and Hutchinson, J.W. (1975), Fully Plastic Crack Problems: The CentreCracked Strip under Plane Strain, International Journal of Solids and Structures, 11, pp. 575–591.
6.7 Kumar, V., German, M.D., and Shih, C.F. (1981). An Engineering Approach for ElasticPlastic Fracture Analysis, Electric Power Research Institute EPRI NP-1931, Project 1287-1,
Topical Report.
6.8 Anderson, T.L. (2004). Fracture Mechanics: Fundamentals and Applications, CRC, Press-Book.
6.9 Sanford, R.J. (2003). Principles of Fracture Mechanics, Prentice Hall, Upper Saddle River.
6.10 Janssen, M., Zuidema, J. & Wanhill, R.J.H. (2004). Fracture Mechanics, Spon Press, Abingdon.
6.11 Gdoutos, E.E. (2005). Fracture Mechanics—An Introduction, Springer, The Netherland.
6.12 Ramesh, K. (2007). e-Book on Engineering Fracture Mechanics, IIT Madras,
URL: http//apm.iitm.ac.in/smlab/kramesh/book_4.htm
Chapter
7
Crack Tip Opening Displacement
If I don't get there headed straight, may be I get there by zig zagging or jumping over
the problem.
Dr Nathan Kline
7.1 INTRODUCTION
Crack tip opening displacement (CTOD) is another parameter suitable to characterize a crack. Unlike
parameters G and K, it can be used for both linear elastic fracture mechanics (LEFM) and elasticplastic fracture mechanics (EPFM). It was formulated by Wells [7.1, 7.2] and Cottrell [7.3] and became
more popular in Europe, at least in the initial stages. In fact, the parameter was formulated about a
decade before it was realized that the J-Integral could be used for EPFM. How is CTOD defined?
The material cannot withstand very high stresses within the plastic zone, and the usual stress
field of the square root singularity no longer exists. However, rigorous analysis is complex and we
would like to explore a simple model. The yielding of the material and resulting rearrangement of
the stresses around the crack tip can be accounted by an effective crack, which is longer than the
actual crack as discussed in Chapter 5. The tip of the effective crack is located inside the plastic
zone (Fig. 7.1). Now, the linear equations of elasticity can be applied to the effective crack. One can
Effective
crack
CTOD
Actual
crack
Fig. 7.1 The effective crack and CTOD
150
Elements of Fracture Mechanics
then visualize that the effective crack has some finite opening at the location of the actual crack tip.
The opening is defined as the crack tip opening displacement as shown in Fig. 7.1.
In reality, there is hardly any opening of the crack tip; only the tip may become more rounded as
the plastic deformation increases. Thus, the definition of the CTOD is based on a model that a
beginner may find a bit difficult to accept. Further, the symbol for a crack tip opening displacement
is still not standardized. Some people use the symbol d, while others call it COD (crack opening
displacement). However, the term ‘COD’ is found ambiguous, because at any location of the crack,
the opening is called a crack opening displacement. Therefore, we will be using CTOD as the
parameter and CTODc, as its critical value.
The CTOD parameter has been found more useful for cracks having large plastic zones. Like the
J-Integral, the CTOD is another approach to deal with elastic-plastic fracture mechanics (EPFM).
However, in the next section, we will explore equivalent relations between K, G and CTOD for
cracks with small plastic zone (LEFM).
7.2 RELATIONSHIP BETWEEN CTOD, KI AND GI FOR SMALL SCALE YIELDING
In this section, we will prove that for small scale yielding in the vicinity of the crack tip
CTOD =
K I2
l Es ys
(7.1)
where l is a constant and its value is close to unity. In fact, its value depends on the plastic zone
and, therefore, l is influenced by the model chosen for determining the plastic zone size. Since
GI = K 2I /E for plane stress, we can express CTOD in terms of the relation
CTOD =
G
.
ls ys
(7.2)
For deriving Eq. (7.1), we will use the expression of crack opening displacement of Mode I,
developed in Chapter 3 [Eq. (3.35)]. Thus, at distance of x1, COD becomes:
COD =
4s
E
2
- x12 for plane stress
aeff
(7.3)
4s (1 - v2 ) 2
aeff - x12 for plane strain
E
where, a eff is the length of the effective crack and can be estimated using Irwin's correction
[Eq. (5.9)] as
rp
aeff = a +
.
(7.4)
2
Substituting aeff in Eq. (7.3), we have for plane stress
COD =
1/2
2
ÈÊ
˘
rp ˆ
Í Á a + ˜ - x12 ˙
2¯
ÍË
˙
Î
˚
For evaluating CTOD, x1 is replaced by a and we obtain
4s
COD =
E
CTOD =
˘
4s È rp
4a + rp ˙
Í
E Î4
˚
(
)
1/2
Crack Tip Opening Displacement
151
For LEFM, rp may be neglected in comparison to 4a and the expression simplifies to
4s
E
where rp is given by Eq. (5.8)
CTOD =
rp =
KI2
2
ps ys
arp
.
KI in this expression is based on aeff. In the case of small scale yielding considered in this section, the
approximate relation is used by determining KI on the actual crack length a. Then, the CTOD is
expressed as
CTOD =
4s
E
Ê s 2p a ˆ
4s 2 a
.
aÁ
˜ =
2
ÁË p s ys ˜¯
Es ys
This expression can be restated by using KI = s p a as
CTOD =
4KI2
p Es ys
.
If the Dugdale approach is used, it can be shown that
CTOD =
KI2
Es ys
GI
=
s ys
.
Thus, the equivalence relations depend upon the model adopted for finding the effective crack
length. Also, the relations depend upon the hardening parameter of work hardening materials.
The value of l in Eq. (7.1) can vary between and p/4 and 2.2. However, direct experimental
methods find that l is closer to unity (through optical methods [7.4] and the metallographic
sectioning of the crack tip filled with a plastic [7.5]).
7.3 EQUIVALENCE BETWEEN CTOD
AND
J
Both the CTOD and the J-Integral characterize elastic-plastic fracture and, therefore, a relation
should exist between them. Consider a crack of actual length a and an effective crack of length aeff
(Fig. 7.2), whose crack tip is located at point B. Figure 7.2 shows CTOD as the distance AC. An
appropriate integration path ABC is chosen for determining the J-Integral which is restated as
J=
Ê
Ú ÁËWdx
G
2
- Ti
∂ui ˆ
ds .
∂x1 ˜¯
Along the integration path, dx2 is negligibly small, making the first term of J negligible. If the
material is considered to be elastic-perfectly plastic, T2 = s ys on the integration path G, then,
J simplifies to
152
Elements of Fracture Mechanics
x2
x1
sys
C
B
A
sys
a
aeff
Finding J-integral along path ABC
Fig. 7.2
Ú
J = – Ti
G
=
∂u2
ds
∂x1
Ú s d (u
ys
BC
2
- u2AB
)
= s ys CTOD
leading to
CTOD =
J
s ys
(7.5)
However, the material within the plastic zone may deform with work hardening, as
characterized by the factor n. The relation is then written in more general form as
CTOD =
J
as ys
(7.6)
where a is a dimensionless factor that depends on n. Test methods to determine CTOD c are
described in Chapter 8.
7.4 CLOSURE
Crack tip opening displacement parameter can characterize materials for both LEFM and EPFM.
This approach is not well-developed to give simple relations between field stresses (or loads) and
CTOD and therefore the parameter is not convenient to predict crack growth in practical situations.
The CTOD approach may be useful to compare toughness of materials. In certain cases, JIc or KIc
may be determined by finding CTODc and then using equivalence relations. However, some
uncertainty exists in using the equivalence relations because they depend on the elastic-plastic
material behavior and the model adopted to find the effective crack length.
Crack Tip Opening Displacement
153
QUESTIONS
1. What are other symbols used in the literature for CTOD?
2. Through a clear sketch, show the distance that depicts CTOD.
3. Why are direct relations not available between CTOD and a load applied on the
component?
4. CTOD was developed about a decade earlier than the application of the J-Integral to
fracture problems began. Still, the CTOD remains unpopular. Why so?
5. The CTOD is a virtual displacement, which is determined through a model. Do you think
that the model is based on a sound approach? Justify your answer.
PROBLEMS
1. Determine GIc, JIc and CTODc for a hardened steel if material properties are: s ys = 700
MPa, E = 207 GPa, KIc = 48 MPa m and n = 0.3.
2. A cylindrical vessel with closed ends is made of steel with yield stress 350 MPa and
modulus 207 GPa. The diameter of the pressure vessel is 1.6 m and the wall-thickness is 18
mm. The critical CTOD is known to be 0.08 mm. Use the small scale yielding model to
determine the maximum pressure the vessel can withstand if a through-the-thickness crack
of length 2a = 20 mm is detected, parallel to the axis of the cylinder.
REFERENCES
7.1 Wells, A.A. (1961). “Unstable Crack Propagation in Metals: Cleavage and Fast Fracture,”
Proceedings of Crack Propagation Symposium, Cranfield, 1, pp. 210–230.
7.2 Wells, A.A. (1963). “Application of Fracture Mechanics at and beyond General Yielding,”
British Welding Journal, 10, pp. 563–570.
7.3 Cottrell, A.H. (1961). “Theoretical Aspects of Radiation Damage and Brittle Fracture in
Steel Pressure Vessels,” Iron Steel Institute Special Report, No. 69, pp 281–296.
7.4 Rooke, D.D. and Bradshaw, F.J. (1969). “A Study of Crack Tip Deformation and a Derivation
of Fracture Energy,” Fracture, 1969, pp. 46–57.
7.5 Bowles, C.Q. (1970). Strain Distribution and Deformation at the Crack Tip in Low Cycle Fatigue,
Army Mat. and Mech. Res. Center, Watertown, Mass. Rept. AMMRC CR 70–73.
Chapter
8
Test Methods
It is said that in old days when an armourer fulfilled an order for iron shirt of mail, he
first put it on himself and his client dealt him a few blows with his sword. If the
craftsman survived, his 'product' was considered satisfactory... But if the armour's work
was not good enough, there was no one to pay the money to.
S. Venestky
8.1 INTRODUCTION
A material is selected to have adequate toughness for a given application. We, therefore avoid
using materials of low toughness like glass, carbides and ceramics for most of the daily structural
applications except in some special cases. One of the main reasons for widespread popularity of
mild steel is that its toughness is very high. Toughness is a material property as yield stress. Proper
test methods should be developed to determine material toughness.
For a long time, people understood toughness of materials intuitively. People could value it but
could not measure it. One crude measure, still sometimes used today, is to find how much a
material elongates before it fails. In the field of engineering, this property is known as elongation.
A material with 10% elongation is much tougher than another material having 2% elongation.
However, the modern fracture mechanics has shown that elongation alone cannot measure the
toughness of a material.
A better estimate of toughness can be realized by finding the area under the stress-strain curve
up to the ultimate tensile strength. Larger area corresponds to higher toughness. Keeping track of
the area under the curve may not be practical for the field engineers. However, the field engineers
adopt a simple method. They keep track of the elongation as well as the ultimate strength of a
material. These two properties give a reasonable estimate of the area under the stress-strain curve.
However, this method does give some feel of the toughness of the material but is not rigorous. It is
used only when modern techniques are not available.
The first method developed for measuring toughness is known as Charpy or Izod notch
sensitivity test. In these tests, a specimen with a notch is impacted by a very large pendulum mass,
Test Methods
155
and the energy absorbed in failing the specimen either through bending or breaking at the notch is
measured. A tougher material absorbs more energy. These test machines became popular rather
quickly and are still used all over. In fact, this kind of notch sensitivity test explained the failure of
ships in the cold waters of the Northern Atlantic Ocean during the Second World War. The steel
used in making the structure of these ships changed its behavior from ductile to brittle below a
certain temperature and, then, even a small flaw in the hull would grow and break the ship into
two parts.
Notch sensitivity impact tests are simple to conduct and not very expensive. However, they are
not based on rigorous analysis. It is difficult to use the results of these tests to predict whether a
crack is likely to grow under given loading conditions. In these tests, the thickness of the specimen
chosen is not based on the toughness of the material and thus considerations of plane stress and
plane strain are not accounted for. In fact, unlike in the rigorous tests of finding KIc and JIc, there is
hardly any consideration in choosing the size of the specimen of Charpy or Izod test based on the
material toughness. Further, notch sensitivity tests are always conducted under dynamic impact
which is generally not the case in many engineering applications. For less ductile materials, the
sharpness of the crack tip is important but rigorous considerations regarding the crack tip
sharpness are not undertaken in the specimen of the Izod or Charpy test. As a postgraduate student
in early seventies at the Brown University, I was surprised to find that the existing impact test
machine was taken out of the test facility area because test methods for J-Integral were being
developed and it would be disgrace to Brown University if some one found out that Charpy tests
were being conducted at the side of J-Integral tests.
However, we will soon realize that evaluation of KIc or JIc is so sophisticated a process that many
field engineers fail to understand the theories and their applications. Even if they understand the
modern test methods, the tests are so expensive that they are out of the reach of many engineering
companies. Consequently, crude methods are still widely used to estimate the toughness of a
material; Charpy and Izod test machines are still popular.
One of the ironies is that research and development resources are mostly used up for theoretical
developments and analysis. These days not much importance and respect is given to experimental
development. Consequently, experimental methods have not been developed and simplified at a
similar pace and whatever have been developed are still out of means of many engineering
companies. To develop new materials, to improve properties of existing materials or to assure
proper material selection of machine components, it is essential to have effective test methods. At
the same time, it is desirable that the test methods are made simple enough for a semi-technical
person to conduct them routinely. Unfortunately, test methods developed so far are quite
sophisticated and beyond the reach of many small and medium size companies.
It is important to emphasize that fracture toughness is a material property. It cannot be derived
from other material properties like modulus, yield stress, etc. However, other properties may
influence toughness of a material. For example, when an alloy steel is given heat treatment to
obtain higher yield stress, its toughness decreases. We cannot calculate or determine toughness just
by knowing the yield stress of the material. A separate experimental test is required to determine
material toughness.
In the previous chapters, four commonly used parameters G, K, J and CTOD have been
discussed. Which parameter is adopted by a designer depends upon the application. K is widely
used for LEFM and J for EPFM. G is generally used for brittle materials and finds its application in
special cases such as interlaminar toughness of fiber composite laminates. CTOD is an alternative
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Elements of Fracture Mechanics
approach to deal with EPFM. Depending upon the extent of usage, we have chosen to present test
methods in the order K, J, G and CTOD in this chapter.
Before we start discussing the details of the various test methods, we would like to comment
that, in general, a crack propagates in a mixed mode, that is, a propagating crack has components
of all three Modes, I, II and III. However, it has been found that Mode I dominates in most of the
real life cases. Therefore, a greater emphasis has been placed on test techniques of mode I. In fact,
test techniques and codes are already available for Mode I, whereas test methods for Mode II and
Mode III have not yet been investigated adequately. Therefore, in this chapter, we will confine our
discussion mostly to Mode I cracks.
8.2 KIc-TEST TECHNIQUE
Experimental determination of the critical stress intensity factor (KIc) is most widely studied and
developed. Code on KIc test was made for the first time in 1965–66 but since then the code has been
revised several times. In this section, we will be discussing the salient features from the ASTM
Code Designation E 399-83 to determine KIc of a material [8.1].
We discussed in Chapter 4 that the critical stress intensity factor of a material depends on the
thickness of the plate. However, for a thick plate it is independent of thickness because the material
in front of crack tip deforms in plane strain. Then, critical stress intensity factor can be treated as
the property of the material. Thus, the experiment should be controlled so as to have its loading in
plane strain only; that is, the plastic zone size in front of the crack tip is quite small in comparison
to the specimen thickness. Then the Linear Elastic Fracture Mechanics (LEFM) can be applied to do
the analysis. The entire body of the specimen is assumed to be deformed elastically and small
strain theory having linear stress-strain relations is invoked.
Before presenting the details of KIc-test, it is summarized as follows. To begin with, KIc of the
specimen is guessed. Then, the specimen is prepared following several dimensional constraints
which are based on the guessed value of KIc. The crack tip is made very sharp with a fatigue growth.
The specimen is pulled in a tensile machine to obtain a relation between the load and the crack
mouth opening displacement. This relation provides the critical load PQ. Accounting for the crack
length and geometry of specimen, the stress intensity factor KQ corresponding to PQ is determined
using LEFM. If KQ satisfies all the constraints on the geometry of the specimen and of fatigue
growth, it becomes KIc. The remaining portion of this section describes the method in detail.
8.2.1 Various Test Specimens
Four kinds of test specimens are shown in Figs 8.1 and 8.2—Compact Tension (CT), Single Edge
Notch Bend (SENB), Arc-shaped Tension (AT) and Disc-shaped Compact Tension (DCT). However,
CT and SENB specimens are the ones widely used. AT specimen is convenient for cylindrical
geometries such as pressure vessels or pipings and DCT specimen is suitable for circular blanks,
round bars, cores, etc. Also, there are two recommended geometries for AT specimen, one with
X/W = 0.5 and another X/W = 0. For introducing a crack in a specimen, a notch is cut by a machine
usually of length 0.45W. The tip of the crack is then sharpened by growing the machined crack
further so as to have crack length a close to 0.5W through a controlled fatigue loading.
Test Methods
Fig. 8.1 (a) CT specimen, and (b) SENB specimen
Fig 8.2 (a) AT specimen, and (b) DCT specimen
157
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Elements of Fracture Mechanics
8.2.2 Constraints on Specimen-Dimensions
Specimen plate thickness B, crack length a and width W must satisfy the following requirements:
ÊK ˆ
B ≥ 2.5 Á Ic ˜
Ë s ys ¯
2
ÊK ˆ
a ≥ 2.5 Á Ic ˜
Ë s ys ¯
2
ÊK ˆ
W ≥ 5.0 Á Ic ˜
Ë s ys ¯
(8.1a)
(8.1b)
2
(8.1c)
where sys is the usual yield stress of the material. Why should these conditions be imposed? The
first condition is required to meet the foremost requirement—the specimen must be loaded in plane
strain. The plastic zone size, based on Irwin’s correction, is close to 0.1 (KIc/sys)2 for plane strain.
When the inequality 8.1(a) is invoked, the thickness of specimen is much larger (about 25 times)
than the plastic zone size, assuring plane strain conditions at the crack tip.
The constraint on the crack length can also be justified. If the specimen is designed to have a
small crack length, the far field stress has to be high to develop large enough SIF at the tip for the
crack to grow. But the large far field stress is undesirable because the analysis would not be
accurate.
The third inequality is also justified because the free surface of lateral faces should be reasonably
away from the crack tip. Although the effect of free surfaces is accounted for in the analysis, the
accuracy improves if the free surfaces are far away from the crack tip.
8.2.3 A Dilemma
Inequalities 8.1 are like the chicken and egg problem. The goal is to have an experimental
determination of KIc but until and unless we know its value, how can we determine the size of the
specimen through the inequalities (8.1a)? One way out is that we take a very thick specimen and
stay on the conservative side. Such approach is not recommended owing to several practical
problems such as difficulty in purchasing a thick plate for the specimen, high machining cost,
specimen handling difficulties and requirement of a high capacity test machine.
For many materials, we already have an idea of the range of KIc prior to the test and one can
design specimen accordingly. If KIc of a new material is to be determined and there is no prior
information about the expected KIc, one would have to guess a value for its KIc , design new
specimens accordingly and perform the test (the details of the test will be presented shortly). The
fracture toughness of the material thus obtained is given a tentative name KQ. Once the value of KQ
is known, specimen dimensions (a, B and W) are evaluated through Inequality 8.1. The test is a
valid test if all the three inequalities are satisfied and then KQ becomes KIc. Otherwise, another
guess on KIc is required and the experiment is conducted again with modified dimensions of the
specimen.
Test Methods
159
8.2.4 Fatigue Crack Growth to Sharpen the Tip
An ideal crack should have a zero radius curvature at the tip because it is the worst case scenario;
the stresses near the tip are highest and the material is weakest against fracture failure. A crack is
generally prepared in a two-step procedure, (i) making a machined slot, and (ii) extending its tip
by fatigue loading. It is worth noting that a cutting tool cannot make a very sharp notch. There is
always a finite radius of curvature of the cutting edge of the tool and also cutting of an engineering
material is associated with some plastic flow. As stated earlier, the slot is machined cut to a length
which is close to 0.45W and then the tip is extended by at least 0.05W with the help of an
appropriate fatigue load.
The tip of the machined slot should be prepared carefully to guide the fatigue-growth. Two
kinds of notches are recommended—(i) V-notch, and (ii) Chevron notch, as shown in Fig. 8.3.
V-notch is relatively simple to prepare but the crack front after the fatigue-growth may get inclined;
the crack front may have a shorter crack length on one face of the specimen in comparison to that
on the opposite face. Specimen having more than 10% difference in the two crack lengths should be
rejected. Further, in some specimen fatigue crack emanating from the V-notch may not even grow
in the plane of the machined slot, and such specimen should be rejected.
Point at mid thickness
Fatigue growth
90°
W/10
W/10
V-Notch
Fig. 8.3 V-notch and Chevron notch
The Chevron notch minimizes the problems of the V-notch but it is relatively difficult to prepare.
The notch is prepared with an angle tool forming four inclined faces near the tip. In fact, a valley is
formed on each side of the specimen. The bottom line of the valley is inclined having one end at the
mid thickness. The Chevron notch works well because the thickness at the crack tip is very small
(tending to be zero) and the fatigue crack nucleates with a small number of load cycles. The
nucleated crack front is guided on the self-similar plane because the thickness is smallest as long as
the crack tip is with in the Chevron notch. Once a crack tip moves for a certain distance on the
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Elements of Fracture Mechanics
self-similar plane, it is expected that it would continue to grow on the same plane when the crack
tip comes out of the notch. If the growing crack deviates away from the plane of the crack, the
specimen should be rejected.
The maximum load of the fatigue loading should not be large because it would then develop a
plastic zone of the significant size. The crack growth will no longer be sharp then. On the other
hand, a low fatigue load is not practical as it would take a large number of load cycles to prepare a
specimen. The following constraint should be followed for fatigue crack growth:
Kf(max) ≥ 0.6 KQ
(8.2)
where Kf(max) corresponds to the maximum value of the fatigue load. Like the previous inequalities,
this inequality also poses difficulties as KQ is not known a priori; Inequality (8.2) should be checked
after KQ is determined. If Kf(max) employed is more than 0.6 KQ, the experiment is rejected and a new
specimen is fatigued at a lower maximum load.
8.2.5 Clip Gauge
A clip gauge measures the Crack Mouth Opening Displacement (CMOD) accurately. It is also
known as COD gauge. The measurement of the displacement should be highly linear and very
sensitive with an accuracy better than 1%.
A typical clip gauge is made of two elastic cantilever strips clamped at the far end of the gauge
with a spacer in between them. Mounting of the clip gauge to the mouth of a crack is designed
carefully. Two kinds of mounting are popular. Figure 8.4 shows the first kind in which free ends of
the strips are prepared to fit into a suitably designed groove on each surface receptor of the
machined notch. The detail of the notched surface shows that a proper knife edge is prepared so
that the end of the gauge strip can be fitted well into the receptor groove. The spacing between the
Strain gauges
Notch
A
Detail A
Specimen
Groove in specimen
with knife edge
Fig. 8.4 (a) Clip gauge mounted on specially prepared receptor grooves,
and (b) details of the receptor groove
Test Methods
161
strips is designed to accommodate substantial bending of the cantilever strips at the time of
mounting the gauge into the notch. When a load is applied to the specimen and the mouth of the
crack opens, the strips strive to become straight. This ensures a good positive contact between the
ends of the gauge-strips and grooved surfaces of the crack-notch. Figure 8.5 shows the other kind
of mounting. The free ends of the thin strips are designed such that they bend outward as shown in
the figure. The bent portions are fastened using screws to the side face of the specimen. Tapped
holes, two on each side of the notch of the specimen, are made on the side face such that they match
with the clear holes in the clip gauge and substantial bending of the strips is achieved at the time of
mounting. Electrical strain gauges are mounted on both sides of each strip as shown in Figs 8.4 and
8.5. A full Wheatstone-Bridge is used to monitor strains in the strips. The gauge is calibrated to
determine distance between the ends of cantilever strips.
Strain gauges
Notch
Fig. 8.5 Clip gauge screwed to the side face of a specimen
8.2.6 Load-Displacement Test
A specimen is loaded to obtain load vs. crack mouth opening displacement (CMOD) using a clip
gauge. CT, AT and DCT specimen are pulled in a tensile machine to record load vs. CMOD relation;
SENB specimen is loaded in a 3-point loading set up with span S = 4W as shown in Fig. 8.1.
Depending upon the material, three types of response are observed between load P and crack
mouth opening displacement as shown in Fig. 8.6. In all three cases P-CMOD curves are linear at
low loads. In the case (a) the curve starts becoming nonlinear with increasing load owing to
growing plastic zone at the crack tip and the stable crack growth. In the case (b) the crack grows
with a pop-in sound and the load drops for a while before it starts increasing again. The case (c) is
the response of an ideal brittle material.
Now we should define a criterion to determine fracture load PQ from these experimental
records. The criterion should guarantee that either the crack tip has already moved by a small
distance or it is definite to move if load is increased by a small amount.
For the case (a) of Fig. 8.6, a 5% secant offset line OS is drawn and its intersection with P-CMOD
curve gives fracture load PQ. The 5% secant line is drawn by having its slope 5% less than the
initial slope of P-CMOD curve. It has been found that 5% secant line corresponds to about 2%
increase in crack length in CT and SENB specimens. Note that 5% is judicially chosen in the code
and is similar to 0.2% offset used in determining yield stress of a material. For finding accurate
value of PQ the P-CMOD curve should be recorded with proper scales such that the initial slope
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Elements of Fracture Mechanics
lies between 0.7 and 1.5. For case (b), the load which initiates pop-in crack growth is taken to be
fracture load PQ. In the case (c) of the brittle growth, the maximum load is PQ.
Pop-in
sound
S
PQ
P
PQ
5% Secant
line
O
CMOD
P
PQ
5% Secant
line
O
(a)
P
O
CMOD
(b)
CMOD
(c)
Fig. 8.6 Load (P) vs. crack mouth opening displacement (CMOD)
8.2.7 Measuring the Crack Length
For accurate measurement of the crack length, it is recommended that the specimen is split into
two halves by letting the crack run all the way. Looking into a split surface, three kinds of surfacezones can be observed—(i) machine cut (ii) fatigue grown and (iii) surface produced through
fracture during the loading. The texture and reflectivity of an incident light from the fatigue-area
are much different from those of the fractured surface and, therefore, three surface zones can be
easily identified. Figure 8.7 shows all the three surface-zones and the commonly observed shape of
the fatigue crack front developed from a Chevron notch. The front of the fatigue grown crack is
generally found to be curved as shown in the figure. The crack length is taken as a = 1/ 3 (a1 + a2 + a3).
The test is rejected if anyone of the three crack lengths differs more than by 5% from a. The test is
also rejected if any part of the crack front of fatigue growth is close to the machined notch within
0.005 a or 1.3 mm.
a1
Fractured
surface
a3
Fatigue
growth
a2
Fig. 8.7 Measurement of crack length
Test Methods
163
8.2.8 Data Analysis
Knowing critical load PQ, crack length a and specimen dimensions, LEFM is invoked to determine
KIc. Specimens are made of finite dimensions and therefore effect of the size must be incorporated
in the data analysis. Appendix 4A gives relevant formulae for each case. For SENB specimen, KI
has the following form:
KI =
PS
f ( a /W )
B W 3/2
(8.3)
and for CT, AT and DCT specimen,
P
(8.4)
f ( a /W ) .
BW 1/2
Form of f (a/W) is quite involved and one may make a mistake in calculating its value for a given a/W.
Table 8.1, therefore, lists its values for SENB, CT and DCT specimen for various a/W.
KI =
TABLE 8.1
Value of f (a/W) for determining KIc
a/W
SENB
f (a/W)
CT
0.450
0.455
0.460
0.465
2.286
2.320
2.354
2.390
8.340
8.458
8.580
8.704
8.705
8.838
8.974
9.112
0.470
0.475
0.480
0.485
2.426
2.463
2.501
2.540
8.830
8.960
9.093
9.230
9.254
9.399
9.547
9.698
0.490
0.495
0.500
0.505
2.580
2.621
2.663
2.705
9.369
9.512
9.659
9.807
9.853
10.01
10.17
10.33
0.510
0.515
0.520
0.525
2.749
2.794
2.840
2.887
9.964
10.12
10.29
10.45
10.51
10.68
10.86
11.04
0.530
0.535
0.540
0.545
0.550
2.936
2.985
3.036
3.089
3.142
10.63
10.82
10.93
11.17
11.36
11.23
11.42
11.62
11.83
12.04
DCT
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Elements of Fracture Mechanics
Equations (8.3) and (8.4) enable us to find KQ corresponding to PQ. If KQ satisfies all dimensional
Inequalities 8.1 and fatigue Inequality 8.2, it becomes the material property KIc. If not, another
guess for KQ is made, next specimen is prepared with modified dimensions, and the entire test is
repeated. Note that it is a trial and error method.
8.2.9 Comments on Plane Strain KIc-Test
For tough materials with low yield stress, thickness required for KIc-test specimen is high and it
would not be practical to do KIc test. Thus, it is difficult to make specimens for tough materials like
mild steel, nuclear reactor steel, commercially available aluminum, etc. In fact, the difficulty
encountered in testing of these materials motivated researchers to formulate J-Integral test which
will be discussed in detail in the next section. However, even if we determine plane strain KIc of
tough material like mild steel, the result may not be useful to us because in most of the practical
applications, sheet thickness falls in the category of plane stress.
Alloy steels are now being increasingly used in engineering applications. Their properties are
altered considerably through heat treatment. One can practically choose almost any yield stress he
wants in his steel alloys. If the yield stress is increased, KIc changes considerably to a lower value.
There are innumerable combinations of KIc and yield stress sys and there are many different kinds
of alloys available in the market. Characterization of KIc for all materials is a voluminous task
because considerable time, care and expenses are required for each test. Consequently, it becomes
difficult for a designer to obtain sufficient data on KIc.
Most of the plates or standard sections such as angle, channel, tube, I-Beam, etc., are made
through rolling with elongated grains in the direction of the rolling. Thus, toughness depends also
on the direction of crack front. For example, a through-the-thickness crack growing in rolling
direction is less tough in comparison to a crack advancing in the direction of the width of the plate.
A KIc test is costly. It needs a well trained person to design specimens. Growing a crack through
a fatigue load is expensive and time consuming. As a result, not many laboratories are equipped to
conduct KIc test. Even after progressing so much in analytical field, a designer faces problems in
selecting a material in the absence of an adequate data bank. In many cases, engineering companies
are not able to afford the KIc test. Designers are forced to somehow manage by taking KIc of a
similar looking material and modifying it with some correction factors based on the yield stress.
8.3 TEST M ETHODS TO DETERMINE JIc
The J-Integral has been developed to account for elastic-plastic behavior of the material in the
plastic zone near the crack tip. Methods to find JIc are quite different from the procedure of finding
KIc. In KIc-test method, specimen is chosen to be of a large thickness so that the plastic zone size
near the crack tip is small and nonlinear stress strain relations in the plastic-zone do not affect the
linear elastic analysis significantly. On the other hand, a specimen of much smaller thickness for
JIc-test can be chosen with a large plastic zone size at the vicinity of the crack tip.
Test Methods
165
8.3.1 Graphical Interpretation
For test methods, the energy interpretation of J-Integral is found to be more convenient. Figure 8.8
shows the energy interpretation for GI for linear elastic material and JI for elastic-plastic material.
Linear load-displacement relations (P vs. u) are shown in Fig. 8.8 (a) for two cracks with length a
and a+Da. It has already been shown in Section 2.6 that area between the two linear lines is equal to
GI BDa where B is the thickness of the plate.
JIBDa
a
GIBDa
P
a
P
a
+
PA
a+
PB
Da
Da
uA
u
u
(a)
uB
uI
(b)
Fig. 8.8 (a) GI through linear P-u relations, and (b) JI through elastic-plastic P-u relations
For elastic-plastic case shown in Fig. 8.8 (b), think in terms of two specimens identical in all
respects except with the slight change in crack length. The area between the two curves
corresponds to JI BDa. The area can be determined from P-u relations in several different ways:
(i) Area under a P-u curve of elastic-plastic deformation (Fig. 8.9) is defined as:
U EP =
Ú Pdu
UEP
P
u
Fig. 8.9 Area under P-u curve (UEP)
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Elements of Fracture Mechanics
Now, consider the case of Fig. 8.8 (b) for two cracks of length a and a+Da. Both specimens
are deformed to the displacement u1. Analogous to definition of G in Chapter 2, we can
express J in terms of UEP. However, for the case of constant u, external work Wext is zero and
expression of JI becomes:
JI = -
1 Ê ∂U EP ˆ
Á
˜
B Ë ∂a ¯ u1
(8.5)
(ii) We find the area under the P-u curve by considering displacement controlled case in which
load decreases from PA to P B as the crack length is increased from a to a + Da as shown in
Fig. 8.8 (b). Then, the area of the thin vertical slice is (–DP)Du. Integrating to obtain the
entire area between the two curves and equating it to B JI D a, we obtain [8.2].
u
1 1 ∂P
JI = - Ú Ê ˆ du
B 0 Ë ∂a ¯ u
(8.6)
(iii) In a load control case, displacement increases from uA to uB as the crack length is increased
by D a, as shown in Fig. 8.8 (b). Again equating the area, we obtain [8.2]:
P
JI =
1 1 Ê ∂u ˆ
dP
B Ú0 Ë ∂a ¯ P
(8.7)
8.3.2 Historical Development
For tough materials such as those used in nuclear reactors, the thickness of KI-test specimen should
be very high so as to meet plane strain requirements and therefore specimen becomes very large as
discussed earlier; determination of KIc is prohibitively expensive. Begley and Landes [8.3]
developed an experimental technique which was based on J-Integral. They made use of Eq. (8.5).
In the method developed, CT specimens of identical geometry with different initial crack length
are prepared, and load-displacement relation for each specimen is obtained on a tensile test
machine. The data are analyzed by finding UEP for a few selected displacements (u1, u2, u3) and then
the curves are plotted between UEP and the crack length for each selected displacement. UEP vs. a
curves are differentiated to make use of Eq. (8.5) for determining JIc .
The technique is quite general and may be adopted even for odd shaped specimen. However,
the technique is not described here in detail because of several practical problems. The technique
requires extensive processing of data; replotting of curves is likely to introduce large errors in some
cases. Then the replotted curves are differentiated. As a general practice, differentiation of
experimentally obtained data is avoided, as far as possible, because a small error may get
amplified. Further, it is difficult to define an appropriate criterion for ‘the onset of crack
propagation’. In spite of all these problems, the work of Begley and Landes played an important
role as a stepping stone for the development of better methods.
8.3.3 Formulation
For more convenient techniques to determine JIc, Eq. (8.6) is further exploited. Consider a SENB
specimen (Fig. 8.10) for which some discussion has been presented in Section 6.5. A crack of length
Test Methods
167
a is introduced in a specimen with thickness B and width W. The uncracked ligament b is about 50%
of width W. The ligament b is subjected to vigorous plastic deformation when load P is applied at
the centre. Deflection of the load point is u and rotation of two segments of the specimen is 2Y.
P
u
b
L
a
W
2y
L
B
Fig. 8.10 SENB specimen with a deep crack
The critical J-Integral, denoted by JIc, is the value of JI which would grow the crack if there is an
incremental increase in JI. However, it is not possible to tell when a crack is at ‘the onset of crack
growth’. We should allow a small crack growth to make sure that ’the onset of crack growth’ has
already been achieved. Based on experience, it has been found that unloading caused by the small
crack growth, does not change the value of JI significantly. Therefore, a small crack growth may be
permitted in experimental determination of JIc.
If area under the P-u record of the bend-specimen is A, we will show that J-Integral is given by [8.4]
2A
(8.8)
Bb
This expression can be proved using dimensional analysis applied to the specimen of Fig. 8.10.
The rotation Y is made of two parts: (i) Yp caused by intense plastic deformation of the uncracked
ligament, and (ii) Ye due to elastic deformation in the specimen. It is worth noting that Yp dominates
specially in the case of a deep crack. The effect of elastic deformation thus may be neglected. Yp
depends on the following parameters: bending moment at the crack plane (M), uncracked ligament
(b), specimen thickness (B), flow stress (s f), modulus (E) and hardening parameter (n). The
dimensionless group M/(s f Bb2) depends upon Yp, s f/E and n. It can be expressed as:
JI =
M
s f Bb 2
(
= f y p , s f / E, n
)
Since Yp = u/L and M = a PL, where a is a factor, the equation is modified to:
168
Elements of Fracture Mechanics
P=
s f Bb2
aL
Êu sf ˆ
fÁ ,
,n
Ë L E ˜¯
Differentiating with crack length a and realizing
(8.9)
∂
∂
we obtain:
=∂a
∂b
2 s f Bb Ê u s f ˆ
Ê ∂P ˆ
ÁË ∂ a ˜¯ = - a L f ÁË L , E , n˜¯
u
(8.10)
Substituting Eq. (8.9) into Eq. (8.10), we have:
Ê ∂P ˆ
2P
ÁË ˜¯ = ∂a u
b
Now, substituting it in Eq. (8.6), we have:
u
JI =
1 1 2P
2A
du =
Ú
B0 b
Bb
(8.11)
In case CT specimen is used, Eq (8.11) is modified to [8.5]
A
g ( ao /W )
Bb
where ao is the initial crack length and
JI =
(
(8.12)
)
g ( ao /W ) = 2 ÈÎ(1 + a ) / 1 + a 2 ˘˚
with a defined as:
a = ÈÎ( 2ao / b)2 + 2 ( 2ao / b )1/2 - ( 2ao / b) + 1˘˚
8.3.4 Details of JIc Test Method
Our objective is to determine JIc corresponding to the initiation of a crack growth. This section
describes the salient features of ASTM Code 813 [8.5]. Similar to KIc-test, two kinds of specimen are
recommended—Single Edge Notch Bend (SENB) and Compact tension (CT). Following constraints
are required on the specimen:
a ≥ 0.5 W
(8.13a)
b ≥ 25 JIc /s f
(8.13b)
B ≥ 25 JIc /s f
(8.13c)
where flow stress sf is defined in terms of yield stress sys and ultimate tensile strength suts as:
(
)
sf = s ys + s uts /2
Test Methods
169
The definition of flow stress accounts for work hardening of the material. Justification of the
constraints on the specimen size will be discussed subsequently. The code requires that the crack
tip should be sharpened by growing a machined notch with a fatigue load. Like in the case of
KIc-test, the specimen size of JIc-test is linked with the value of JIc which is not known a priori.
Therefore, specimen size is based on a guessed JIc and the resulting value of J-Integral is called JQ.
If JQ satisfies Inequalities (8.13), it is a valid test, otherwise specimen dimensions should be
modified for the next test.
For evaluating JIc , several specimens of identical size and almost of same initial crack length are
prepared. Each specimen is loaded in a tensile machine to obtain load vs. displacement (P – u)
curve as shown in (Fig. 8.9). The crack is allowed to extend by a small length and the crack
extension is varied from specimen to specimen. The crack extension should be measured
accurately. Before the specimen is broken apart, the extension should be marked or frozen. This can
be achieved through several techniques. The specimen can be heat-tinted. Heat tinting for steel is
carried out for 10 minutes at 800oC. After the heat tinting, which chahnges the color of the cracked
surface due to oxidation, the specimen is broken apart for measuring Da. Another technique is to
use a dye penetrant which would color the cracked surfaces. Dye penetrants are easily available in
the market as they are widely used to detect surface cracks.
Nature of the cracked surface of a specimen is shown in Fig. 8.11. The crack front is generally
curved as shown and it is important to measure D a very accurately. In a thick specimen, it is
recommended that the crack length should be determined by taking average of nine measurements
across the thickness of the specimen.
Surface of broken
specimen
Fatigue
growth
Crack
extension Da
Machined
slot
Fig. 8.11 Crack extension
JI is determined from each specimen using Eq. (8.8) and is plotted against crack extension as
shown in (Fig. 8.12). A straight line, known as R-line, is best fitted through the JI points. A
(
)
blunting line, J I = s ys + s uts Da , is drawn whose intersection with R-line gives JQ as shown in the
figure. It is now worthwhile to justify the form of blunting line and the constraints on the
specimen. Because of the intense plastic deformation the crack-tip blunts before stable crack
170
Elements of Fracture Mechanics
growth initiates. The blunting can be considered as a small crack growth as shown in Fig. 8.13.
For estimating the small crack growth, the blunt crack can be modeled as a semicircle with its
radius CTOD/2. Since CTOD = JI /s f, the small crack growth due to blunting is JI /2sf and
therefore the ‘on-set of crack propagation’ corresponds to:
JI = 2s f D a
(
)
= s ys + s uts D a
(8.14)
Blunting line
R-line
JI
JQ
(sys + suts)
1
Da
Fig. 8.12 Determining JQ through R-line and blunting line
Blunting
Da = JI /(2sf)
Crack
extension
Fig. 8.13 Blunting of crack tip and crack extension
Test Methods
171
Thus, stable crack growth is beyond the blunting line. The nearest experimental point can never
be at the blunting line owing to practical considerations; there has to be a small but finite crack
extension for accurate measurements. Since R-line is extrapolated to find JQ, it is prone to data
processing error. To minimize the error, the nearest experimental point should be close to the
intersection of blunting line and R-line.
JIc measurements require that the plate thickness B should be much larger than the size of the
intense plastic deformation which is given by JI/2s f . Thus, Inequality (8.13c) assures that B is
much bigger than the intense plastic deformation. Similarly, Inequality (8.13b) assures that the
uncracked ligament is much larger than the intense plastic zone. Inequality (8.13a) is required to
enhance plastic deformation in front of the crack tip .
8.3.5 Comments on JIc-Test
JIc corresponds to the onset of crack extension because stable growth is not admissible as it involves
unloading. In most of the real life materials, a crack goes through a fairly large stable growth before
it becomes unstable. Thus, JIc is conservative in comparison to KIc which admits some stable crack
growth. However, in sensitive applications, like in a nuclear reactor, even a small extension of a
crack is considered as failure and JIc is found to be more suitable.
Conducting JIc-tests is expensive because several specimens should be used to draw the R-line.
Further each experiment should be prepared carefully with the expensive fatigue crack growth.
Measurement of crack extension, D a, involves further expense in marking the crack front (e.g., heat
tinting) and making accurate measurements. Consequently, data on JIc are not readily available.
Further, a large but natural scatter in JIc discussed in Secs 6.5 and 6.6, discourages tabulation of
material properties.
Now, we would explore how JIc-test helps in reducing the size of a test specimen. Minimum
recommended thickness BK of KIc-test specimen is:
Ê K2 ˆ
BK = 2.5 Á 2Ic ˜
ÁË s ys ˜¯
and minimum thickness BJ for JIc-test specimen is:
(
BJ = 25 J Ic /s f
)
For the sake of comparing the thickness in the two kinds of tests, we take s f = s ys to have:
( ) /(10 J s )
BK / BJ ª KIc
2
Ic ys
Taking JIc ª KI2c / E
BK / BJ ª
E
10s ys
For most of the engineering materials, we know that modulus is much larger than yield stress.
For reactor steel with E close to 207 ¥ 103 MPa and yield stress of 350 MPa, the ratio of thickness is
172
Elements of Fracture Mechanics
about 60—a dramatic reduction in specimen size. JIc-test enables us to determine toughness of a
ductile material accurately by using a reasonable sized specimen. This has made the test popular
within a short duration especially with critical industries like nuclear plants.
8.4 TEST M ETHODS TO D ETERMINE G Ic
AND
GIIc
Critical energy release rate of commonly used isotropic materials (metal, plastics) is not usually
determined. However, in certain special cases, parameter G is found to be more appropriate. A
laminate of fiber composite material is known to possess poor interlaminar toughness. Similarly,
when two sheets made of metals are adhesively bonded together, the bond is usually weak against
interlaminar crack growth. Thus appropriate techniques should be developed to measure
interlaminar toughness of small magnitude. It is worth noting that stresses around the tip of an
interlaminar crack are low and anelastic deformation is small. Therefore, LEFM is appropriate for
such cases and determination of GIc is usually preferred.
In this section, test techniques for determination of critical energy release rate of interlaminar
crack are discussed for Mode I and Mode II.
8.4.1 Determination of Interlaminar GIc
A Double Cantilever Beam (DCB) specimen of thickness B and height of each cantilever h is
employed to determine critical energy release rate. The end of each cantilever is pulled in a tensile
test machine and the loads are applied through hinges as shown in Fig. 8.14. The hinges release any
bending moment which may otherwise get developed due to the rotation of the cantilever end. The
other end of the specimen tends to move downwards due to its dead weight. This inclination is
eliminated by a small counter weight tied to specimen through a thread as shown in the figure.
P
u
DCB specimen
2h
a
Small counter
weight
Fig. 8.14 The cantilevers of a DCB specimen pulled through hinges and a counter
weight to balance the dead weight of the specimen
Test Methods
173
The overview of the test techniques is as follows: (i) experiments are conducted to determine
compliance of the DCB specimen for several crack lengths, (ii) critical load is determined for each
crack length, and (iii) GIc is obtained with a suitable data reduction scheme [8.6].
The DCB specimen is pulled in a tensile test machine on displacement control with a low pulling
speed and the crack is allowed to grow by a small distance usually in the range of 5-15 mm. The
machine is stopped for some time (till the crack tip becomes stationary) and then the length of the
extended crack is measured. The load carried by the specimen decreases from point A to point D (Fig.
8.15) because the stiffness of the specimen decreases with crack growth. The slope of the u - P curve
up to point A gives the compliance; the load at point A is the critical load Pc for the crack length.
10
87
.2
6
80
.7
Displacement u, mm
8
4
.8
62
7
2
51.
D
40.4
A
a = 30.1 mm
0
0
100
200
Load P, N
300
Fig. 8.15 Loading-unloading cycles to determine compliance
and Pc for several crack length
Now, our aim is to find compliance for the increased crack length. To achieve it, the load on
the specimen is decreased by moving the jaw of the tensile machine in the opposite direction till
the load becomes zero. The unloading curve is usually not linear and is ignored. The specimen is
loaded again to have further crack growth by a small distance. The loading curve gives the
compliance of the increased crack length with critical load at point D. Compliance and critical
load are determined by repeating the process several times as shown in Fig. 8.15. GI of DCB
specimen is given by (Sec. 2.7)
GI =
P 2 a2
EIB
(8.15)
174
Elements of Fracture Mechanics
It is worth noting that modulus E and moment of inertia I may not be known to the investigator
because composite material is not a standard material like steel. The modulus may vary from one
laminate to another. We would not determine E and I through separate tests. Further, composite
materials or bonded sheets do remain elastic during interlaminar growth and therefore we can use
elastic solution for the deformation of DCB specimen. Usually, one specimen grip of the tensile
machine remains stationary; the force and the displacement of the other specimen grip are recorded
accurately. Accounting for the deflection of both cantilevers, displacement u is given by:
u=
2 Pa3
3 EI
and then the compliance C is given by:
u 2 a3
=
P 3 EI
The equation is expressed as:
C=
C = A1 a3
(8.16)
where A1 is a constant. In fact, A1 determines the flexural rigidity EI with the relation:
2
(8.17)
3EI
In case flexural rigidity is well known, one need not evaluate C vs. a relation. However, we would
obtain flexural rigidity directly from loading and unloading cycles of the experiment for most of
the cases. When C and a are plotted on a log-log scale, Eq. (8.16) should give a straight line with
slope 3. Thus, we fit a straight line to the experimental data points with slope 3 and ln(A1) is read
from the graph (Fig. 8.16). In fact, regression analysis can be used to evaluate A1 without plotting
points on a graph paper. The details of the regression analysis are given in Appendix 8A.
A1 =
ln (a)
0
1
2
3
4
5
0
–5
ln (C)
ln (A1)
3
– 10
1
– 15
– 20
Fig. 8.16 Fitting a straight line of slope 3 to determine A1
Test Methods
175
Corresponding to critical load Pc, Eq. (8.15) gives critical release energy rate as
GIc =
P 2c a 2
EIB
Note that GIc is a material property and Pc and a are the only two variables in the above expression.
Defining the product of these variables as:
A2 = Pc a
we obtain:
GIc =
A 22
EIB
(8.18)
In order to evaluate A2 we note that
A2
(8.19)
a
Figure 8.17 shows the plotted values of Pc and a on a log-log scale with best fitted line of slope-1;
the intercept of the line with ln(Pc) axis yields the value of A2. Substituting for EI from Eq. (8.17)
into Eq. (8.18) we obtain:
Pc =
GIc =
3 A 1 A 22
B
2
(8.20)
10
6
–1
ln (A2)
ln (Pc)
8
4
1
2
0
0
1
2
3
4
5
ln (a)
Fig. 8.17 Fitting a straight line of slope –1 to determine A2
Example 8.1 Consider a DCB specimen made of fiber composite laminate whose cantilevers are
pulled in a tensile machine for several crack lengths as shown in Fig. 8.15.
Determine GIc if the specimen thickness is B = 25 mm.
176
Elements of Fracture Mechanics
Solution: Crack length a, compliance C and critical load Pc are determined in Fig. 8.15 and
tabulated in Table 8.2 along with the logarithmic values. Plots of ln (C) vs ln(a) and ln(Pc) vs ln(a)
are shown in Fig. 8.18 along with the best fit lines of slopes 3 and -1. Through regression analysis
(Appendix 8A), we have:
6.0
ln(Pc)
ln(C)
–3
3
–5
5.5
–1
5.0
1
1
–7
4.5
5
4
3
5
4
3
ln(a)
ln(a)
Fig. 8.18 (a) Variation of ln(C) with ln(a), and (b) variation of ln(Pc) with ln(a)
Table 8.2 Logarithmic values of a, C and Pc of Example 8.1
S.No.
1
2
3
4
5
6
a
(mm)
C
(mm/N)
30.1
3.33 ¥ 10–3
40.4
8.14 ¥ 10–3
51.7
14.12 ¥ 10–3
62.8
25.19 ¥ 10–3
80.7
32.40 ¥ 10–3
87.2
55.90 ¥ 10–3
Sum of column (S)
Pc
(N)
ln(a)
ln(C)
ln(Pc)
305.7
266.7
219.5
183.0
166.0
146.0
3.40
3.70
3.95
4.14
4.39
4.47
24.05
– 5.70
– 4.81
– 4.26
– 3.68
– 3.43
– 2.88
– 24.76
5.72
5.59
5.39
5.21
5.11
4.98
32.00
1
1
ÈÎ-3 S ln( a) + S ln(C)˘˚ = ÎÈ-3 ¥ ( 24.05) - 24.76˘˚ = -16.15
n
6
A1 = 96.86 ¥ 10–9 1/N(mm)2
ln (A1) =
yielding
and
yielding
1
1
[ S ln ( a) + S ln (Pc )] = [24.05 + 32] = 9.34
6
n
A2 = 11.38 ¥ 103 Nmm
ln (A2) =
Substituting A1 and A2 in Eq. (8.20), we have
2
3
-9
3 A1 A22 3 ( 96.86 ¥ 10 /Nmm ) ¥ (11.38 ¥ 10 Nmm)
= ¥
GIc =
2 B
2
25 mm
= 0.753 N/mm = 753 J /m2
2
Test Methods
177
8.4.2 Determination of Interlaminar GIIc
Interlaminar critical energy release rate for Mode II is determined by employing an end notched
flexure specimen shown in Fig. 8.19. The edge crack is on the midplane where the shear stress is the
highest. The initial crack length is approximately equal to 0.5 L but it should not exceed 0.69 L. The
specimen is loaded in a 3-point bend fixture as shown in Fig. 8.19. For this system it has been
shown [8.7] that:
GII =
9P 2 Ca 2
(
2B 2L 3 + 3 a 3
)
(8.21)
Fig. 8.19 Edge-notched flexure specimen loaded in 3-point bend fixture
where C is the compliance. The compliance can be found using the beam theory as:
C=
2 L 3+ 3 a 3
8 EBh 3
However, C is not evaluated using this formula due to uncertainty in the modulus of a laminate. It
is determined directly from the P-u record.
We would now check the stability of crack growth. Substituting C in Eq. (8.21) and differentiating
the resulting expression with respect to a we have:
dG II
9P 2 a
=
da
8 EB 2 h 3
178
Elements of Fracture Mechanics
Note that dGII /da is always positive in load control testing and once the crack becomes critical, its
growth is unstable. The case of displacement control is more interesting. Substitution of P = u/C in
Eq. (8.21) yields:
GII =
9u2 a 2
(
2CB 2L3 + 3a 3
)
Substituting the expression of C in this equation and differentiating with respect to a, we obtain:
dG II
72 u 2 Eh 3 a
=
da
( 2L 3 + 3 a 3 ) 2
È
˘
9 a3
1
Í
3
3˙
Î 2 L +3a ˚
For unstable crack growth, dGII/da should be positive, giving:
a£
L
= 0.693 L
( 3)1/3
The unstable crack growth for a < 0.69 L is desirable because it gives a sharp value of critical load Pc
as shown in Fig. 8.20 for Mode II. In fact, the crack acquires high speed in a short duration at the
critical load Pc. Knowing Pc and compliance C from Fig. 8.20 we can determine GIIc using Eq. (8.21).
1500
Load P, N
Pc
1000
500
1
C
0
0
1
2
3
Displacement u, mm
Fig. 8.20 P-u record to determine compliance and critical load
Example 8.2 An end-notched specimen is made of height 2h = 2.8 mm, initial crack length a = 26 mm,
total length between the support 2L = 100 mm and thickness B = 25 mm. The load-displacement
relation under 3-point loading is shown in Fig. 8.20. Determine the critical energy release rate GIIc.
Solution: From Fig. 8.20 compliance C is evaluated in the linear portion of P-u record as:
C = 1.67 ¥ 10–3 mm/N = 1.67 ¥ 10–6 m/N
Test Methods
179
and the critical load is:
Pc = 1361 N
Then, we have:
GIIc =
9Pc2 Ca2
(
2B 2L3 + 3a 3
)
9 ¥ (1361 N) ¥ (1.67 ¥ 10-6 m/N) ¥ (0.026 m)
2
2
= 1243 J /m2
3
3
È
˘
2 ¥ (0.025 m) ¥ Î 2 ¥ ( 0.05 m) + 3 ¥ (0.026 m) ˚
To sum up, in this section experimental techniques have been presented for determining
interlaminar toughness of fiber composite laminates or two strips bonded together for both Mode I
and Mode II cases. Energy release rate parameter has been found to be the most convenient. GIc is
determined by employing a DCB specimen with repeated load cycles for different crack lengths.
The relations between compliance and crack length, and critical load and crack length then yield
GIc. To find GIIc, an end notched specimen is used which is loaded under a 3-point bend test. The
compliance and the critical load for unstable crack growth determine GIIc.
=
8.5 DETERMINATION
OF
CRITICAL CTOD
The definition and the formulation of crack tip opening displacement are presented in Chapter 7.
In this section, an experimental technique to determine the critical crack tip opening displacement
(CTODc or dc) will be discussed.
We would present the salient features of the test technique discussed in the British Standard BS
5762 [8.8]. The code recommends the use of SENB (single-edge-notched bend) specimen to be
loaded under a 3-point bend fixture. The geometry of the specimen is similar to one adopted for
KIc-test. However, the V-notch, rather than the Chevron notch, is recommended. The specimen
should be pre-fatigued as prescribed for KIc-test. Crack mouth opening displacement (CMOD) is
measured when load P is applied at the centre of the SENB specimen.
The general criterion is that load and CMOD are determined when the crack is at the onset of an
unstable crack growth. With an appropriate data reduction scheme, CTODc is determined.
However, the character of P-CMOD record varies from material to material. Often, it is difficult to
determine when exactly the unstable growth initiates.
We will now discuss the various kinds of experimental measurement of P-CMOD relations. The
simplest case deals with initiation of unstable crack growth with pop-in. In Fig 8.21 (a), there is no
stable crack growth prior to pop-in whereas stable growth occurs in the case (b) prior to pop-in.
The load at which pop-in initiates is taken as the critical load. For this critical load, plastic
component up of measured CMOD is found by drawing a line parallel to the initial P-CMOD record
as shown in Fig. 8.21.
We would now consider the cases when pop-in response is not observed. Figure 8.22 (a) deals
with the case of no stable growth and unstable crack growth is initiated at the maximum load Pmax.
In case (b), stable growth occurs prior to unstable crack growth and unstable crack is initiated at
the maximum load Pmax. In case (c), stable growth occurs but unstable growth is initiated prior to
the maximum load. However, it is difficult to identify the load that initiates the unstable crack
180
Elements of Fracture Mechanics
growth in routine testing. Therefore, maximum load is taken to be critical load. Thus, in all three
cases of Fig. 8.22, maximum load is treated as the critical load of unstable crack growth. Plastic
component up is also shown for these cases in the figure.
Pop-in
Pop-in
P
P
up
up
CMOD
CMOD
(b)
(a)
Fig. 8.21 (a) P-CMOD record with pop-in and no stable crack, and (b) with stable
growth before pop-in
Pmax
Pmax
Pmax
P
P
up
CMOD
P
up
(a)
CMOD
(b)
up
CMOD
(c)
Fig. 8.22 P-CMOD record for (a) no stable growth, (b) stable growth prior to unstable growth
initiating at Pmax and (c) stable growth but unstable growth is initiated before Pmax
The CTODc is made of elastic and plastic parts given as:
CTODc = (CTODc )e + (CTODc )p
Test Methods
181
The elastic part (CTODc)e is calculated by finding KIc from P-CMOD records using secant line as
described in Sec. 8.2. Then, the Dugdale model is invoked to find (CTODc)e. For plane strain, the
result of Dugdale model is modified by a factor of ½ to account for constraints on plastic
deformation. Thus, the elastic portion of (CTODc)e is given by the expression:
(CTOD c )e
=
(
K 2Ic 1 - v 2
)
(8.22)
2 s ys E
The plastic portion of (CTODc)p is determined by assuming that the uncracked ligament works
like a plastic hinge with its centre at a distance rb from the crack tip as shown in Fig. 8.23(a).
Knowing up, rb and a, we determine (CTOD c)p through simple analysis of similar triangles
[Fig. 8.23(b)] as:
(CTODc )p
=
up rb
(8.23)
a + rb
Centre of plastic hinge
rb
b
rb
W
(CTOD)p
a
a
up
(a)
(b)
Fig. 8.23 (a) Plastic hinge with its centre at distance rb from the crack tip and (b) obtaining
(CTODc)p through similar triangles
The factor r is determined through experimentation, and it is found to lie between 0.33 and 0.48.
However, experimental determination of r is difficult on a routine test basis; therefore, a nominal
value of r = 0.4 is used for standard testing of CTODc giving
(CTODc)p =
0.4up (W - a)
0.6a + 0.4W
(8.24)
Example 8.3 A single-edge-cracked bend specimen is loaded in a 3-point bend fixture (Fig. 8.1)
and variation of centre load with CMOD is recorded. The span of the bend fixture is s =200 mm for
the specimen of width W = 50 mm, thickness B = 25 mm, and crack length a = 25 mm. P-CMOD
record of the test is shown in Fig. 8.24. Determine CTOD c following British Standard BS 5762 if
E = 207 GPa, v = 0.29, and sys = 600 MPa.
Solution: S = 0.2 m, W = 0.05 m, B = 0.025 m, a = 0.025 m, E = 207 ¥ 103 MPa, v = 0.29. From
Fig. 8.24,
Pc = 26600 N (through 5% secant line)
182
Elements of Fracture Mechanics
Pmax
30
Pc
P, kN
20
5% secant line
10
up
0
1
2
3
4
CMOD, mm
Fig. 8.24 P-CMOD record. For determining (CTODc)e , 5% secant line is drawn to find KIc.
For evaluating (CTODc)p up corresponding to Pmax is found
Pmax = 29800 N
up = 1.14 mm
a = a/W = 0.5
Through Table 8.1 we find f(a) as:
f(a) = 2.663
KIc = f (a )
PcS
BW
3/2
=
2.663 ¥ ( 26600 N) ( 0.2 m)
(0.025 m) ¥ (0.05 m)3/2
= 50.7 MPa m.
Substituting in Eq. (8.22), we have:
(CTODc)e =
(
KI2c 1 - v 2
2s ys E
) = ( 50.7 MPa m ) (1 - 0.29 )
2 ( 600 MPa) ¥ ( 207 ¥ 10 MPa)
2
2
3
= 0.0095 mm
Through Eq. (8.24), we have:
(CTODc )p
=
0.4up (W - a)
0.6 a + 0.4 W
=
0.4 ¥ (1.14 mm ) ¥ ( 0.05 m - 0.025 m )
0.6 ¥ 0.025 m + (0.4 ¥ 0.05 m )
= 0.325 mm
The overall CTODc becomes:
CTODc = 0.0095 + 0.325 = 0.335 mm
Note that the elastic part of CTOD is negligible.
Test Methods
183
8.6 CLOSURE
In this chapter test techniques are presented with reasonable details for finding KIc, JIc , GIc , GIIc,
and CTODc. Out of all these, KIc is the most widely used and its values are listed in literature for
commonly used engineering materials. Determination of KIc is a relatively simple technique but a
good quality CMOD gauge is required. Designs based on plane strain KIc are too conservative for
high toughness materials. Also, determination of KIc for these materials is very difficult and
expensive due to large thickness of the specimen. Therefore, EPFM should be used for which JIc or
CTODc are required. In comparison to CTODc it is found that JIc is more convenient for designers.
However, methods to find JIc are still complex, expensive and time consuming. It is hoped that
simpler test-technique will be developed in future. GIc and GIIc are found convenient for
determining interlaminar toughness of composite materials. The test method for finding GIc is
better developed and more widely used.
APPENDIX 8A
Regression Analysis to Fit a Line of Known Slope
Data analysis for finding GIc requires fitting a straight line of a known slope to experimentally
recorded data. The line is y = mx + c for which slope m is known and c is to be determined. The line
is best fitted so as to minimize sum of square of the distance from data points to the line. Consider
a point (xi, yi) and if the distance of the point from the line is di, the sum D of the square of the
distance for n points is:
n
D=
Â
i =1
For minimum D,
di2
=Â
(mxi - yi + c)2
m2 + 1
dD
=0
dc
leading to
c=
For m = 3,
c=
and for m = –1,
c=
(-Smxi + Syi )
n
(-3S xi + Sy i )
n
( Â xi + Â yi )
n
QUESTIONS
1. Why are results of Charpy or Izod impact tests not useful in predicting loads that would
grow an existing crack in a component with known geometry and boundary conditions?
2. Why are Charpy or Izod tests still popular in industries?
3. Why are the dimensions of specimen for plane strain KIc-test based on material toughness
KIc?
Elements of Fracture Mechanics
184
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
Why is it difficult to find plane strain KIc of ductile materials?
Why is it required to extend a machined notch by fatigue for KIc-test?
There is a constraint on the maximum stress of fatigue load. Why so?
Why is Chevron notch better than V-notch for KIc-test?
In comparison to V-notched specimen, it is observed that fatigue crack initiation occurs at
less number of cycles in Chevron notched specimen. Why so?
What does 5% secant line assure in KIc-test?
Why is texture of fatigue growth different from the texture of fractured surface?
Use of KIc by a designer is on conservative side. Comment on the statement.
KIc can be determined through one specimen, whereas to determine JIc several specimens
are required. Why so?
Why do dimensions of specimens for JIc-test depend on JIc?
How do we implement the requirement of ‘onset of crack initiation’ in JIc-test?
Determination of critical SIF for plane stress is less rigorous because plastic zone size is
large, and linear elastic analysis deviates considerably from the actual stresses and strains.
However, JIc can be obtained quite accurately for plane stress cases. Why so?
GIc-test is found to be best suited for experimental determination of low toughness of certain
cases such as interlaminar crack growth in fiber composite laminates. Why is KIc-test not
suitable?
In GIc-test of interlaminar cracks, flexural rigidity (EI) is evaluated through the experiment
on a DCB specimen and not through a separate experiment such as a tensile test. State the
reasons.
In processing the data of GIc-test, how are the theoretical results exploited in the regression
analysis?
Why do we like to have unstable crack growth in interlaminar GIIc-test?
What are important requirements of a clip gauge?
PROBLEMS
1. The machined Chevron notch of SENB specimen is extended by a fatigue load with Pmax = 40
kN and Pmin = 5 kN. The dimensions of the specimen are:
Thickness B = 25 mm
Width W = 50 mm
Span S = 100 mm
Average crack length a = 24.1 mm.
The critical load determined through 5 % secant line is 80 kN and the yield stress of the
material is 800 MPa. Determine KQ and check whether the dimensions of the specimen
and fatigue load were proper.
2. A compact test specimen is employed to find plane strain KIc. The dimensions of the
specimen are:
Thickness B = 30 mm
Width W = 100 mm
Average crack length a = 49.8 mm
Test Methods
185
The experimental record of P-CMOD is shown in Fig. 8.25. Determine KQ and check whether
the experiment is valid if the yield stress of the specimen material is 600 MPa.
120
Load P, kN
100
80
60
40
20
0
0
0.4
0.5
1.2
1.6
CMOD, mm
Fig. 8.25 The figure of Problem 2
3. Four SENB specimens of an unknown steel material were prepared to find JIc of the following
dimensions:
Thickness B = 20 mm
Width W = 40 mm
Span S = 160 mm
The machined notch is fatigue grown till overall crack length is close to 0.5 W.
Area (A) under the load displacement curve for a small crack growth is determined. The
details of uncracked ligament b, crack extension D a, and area A are:
Expt. No
b
(mm)
Da
(mm)
A
(Nm)
1
2
3
4
19.2
19.6
18.7
20.1
0.36
0.82
1.22
2.70
22.46
25.08
26.93
38.19
The yield stress and ultimate tensile strength of the material are 350 MPa and 550 MPa
respectively. Determine JIc and then find whether dimensions of the specimen were valid
for finding JIc.
4. A bend specimen is loaded under four-point bending as shown in Fig. 8.26. Develop the
expression for J if load displacement (P – u) record is known.
Elements of Fracture Mechanics
Fig. 8.26 The figure of Problem 4
5. A DCB specimen of thickness B = 30 mm and made of a composite material is pulled in a
tensile test machine under displacement control test to find interlaminar toughness. Figure
8.27 shows the load-displacement record of loading and unloading cycles. The crack length
14
12
Displacement, mm
186
67
10
58
8
50
6
4
42
2
a = 32
mm
0
0
20
40
60
80
Load, N
Fig. 8.27 The figure of Problem 5
100
Test Methods
187
of each loading is also shown in the figure. Determine critical energy release rate of
interlaminar crack. Also, determine GIc through individual triangles; find the average and
then compare it with GIc obtained by evaluation of AI and A2.
6. An end-notched flexure specimen of a polymer composite is of length 2L = 160 mm,
thickness B = 22 mm and overall height 2h = 3.2 mm (Fig. 8.19). A precrack of length a = 40
mm is introduced at a midplane of the specimen. Estimate the critical load if the estimated
GIIc is 1200 J/m2 and modulus in the direction of specimen length is 110 GPa.
REFERENCES
8.1 Annual book of ASTM Standards (1987). Metals Test Methods and Analytical Procedures,
ASTM Publications, Philadelphia, 03.01, pp. 522–557.
8.2 Hertzberg, R.W (1989). Deformation and Fracture Mechanics of Engineering Materials, John
Wiley and Sons, New York.
8.3 Begley, J.A. and Landes, J.D. (1972). The J-integral as a Fracture Criterion, Fracture Toughness
Part II, ASTM STP 514, American Society for Testing and Materials, Philadelphia, pp. 1–20.
8.4 Broek, D. (1982). Elementary Engineering Fracture Mechanics, Martinus Nijhoff Publishers,
The Hague.
8.5 Annual Book of ASTM Standards (1986). Metals Test Methods and Analytical Procedures,
ASTM Publications, Philadelphia, 03.01, pp. 772–786.
8.6 Carlsson, L.A. and Pipes, R.B. (1987). Experimental Characterization of Advanced Composite
Materials, Prentice-Hall Inc., New Jersey.
8.7 Carlsson, L.A., Gillespie Jr., J.W. and Pipes, R.B. (1986). On the Design and Analysis of the
End Notched Flexure (ENF) Specimen for Mode II Testing, Journal of Composite Materials,
20, pp. 594–604.
8.8 BS 5762 (1979). Methods for Crack Opening Displacement COD Testing, British Standards
Institution, London.
8.9 Ramesh, K. (2007). e-Book on Engineering Fracture Mechanics, IIT Madras, 2007.
URL: http//apm.iitm.ac.in/smlab/kramesh/book_4_htm
Chapter
9
Fatigue Failure and
Environment-assisted Fracture
Fatigue is a process that is capable of selecting the weakest link in strength
V. Finkel
9.1 INTRODUCTION
The performance of a machine or structural component is usually tested in a laboratory where
conditions are controlled. In real life conditions, the component is subjected to several harsh
conditions. Fluctuating loads, known as fatigue loads, quite often cause failure of components. In
addition, the combined effect of stresses and a corrosive environment may cause a premature
failure of a component. It is now well-known that both of them cause the growth of subcritical
cracks to their critical length, followed by the rapid crack growth and eventual failure of the
component. It is difficult to protect the component from various fluctuating loads as we live with
many machines which either rotate or have back and forth motions. Also it is difficult to control the
service environment of a component. Even abundant materials like water or its vapors, various
salts, oils, edible items are known to cause the growth of subcritical cracks to their critical length.
Thus the designer of a component should identify its service conditions and if required, he should
carry out an appropriate analysis to avoid its premature failure.
In this chapter, various aspects of fatigue failure and environment-assisted fracture are
presented. In addition, a brief discussion on the combined effect of a fluctuating load and a
corrosive environment is included.
9.2 FATIGUE FAILURE
A component, which fails through yielding at a high constant (unfluctuating) load, may fail under a
substantially smaller fatigue load. To consider an analogy, our muscles are not strong enough to
make a tree fall. We can chip off the trunk bit by bit with an axe until the groove becomes so large that
the tree can be pulled down easily with a rope. The process is slow but repeated action makes it fall.
Fatigue Failure and Environment-assisted Fracture
189
Innumerable examples of fatigue loads can be observed in our nearby environment. Many
machines use rotating shafts. Shafts are designed to carry torque, but lateral loads, which generate
bending moments, cannot be avoided. Consequently, a fiber on the shaft surface which is aligned
parallel to the axis is subjected to tensile and compressive stresses in every rotation. Consequently,
the axles of railway wagons, automobiles, and gear boxes are subjected to fluctuating loads. Failure
of a locomotive axle should be avoided as it can stall the entire train suddenly with the wagons
tumbling and climbing over each other. One serious accident took place only two kilometers away
from IIT Kanpur as late as in 1980s. One axle of a goods train failed and, consequently, most of the
wagons were derailed and deshaped. It was a huge financial loss only due to fatigue failure in one
of the axles of the train.
The wings of an airplane are subjected to fluctuating loads of wind gusts during the flights and
therefore, the wings are carefully designed. In fact, before the flight test of a newly developed
airplane, the entire wing is experimentally tested to fluctuating wind loads, simulated in the
laboratory, till it fails. In another example, the fuselage of a plane is subjected to one cycle of
pressure per flight because at high altitude the air pressure is increased inside to make it
comfortable for the passengers. The fuselage of the ill-fated commercial Comet Jet airplanes failed
in 1950s owing to a fatigue crack growth nucleated near an opening in its fuselage.
In sophisticated way of modern living, many automated machines and gadgets have been
developed which involve some kind of rotation, reciprocating motions or vibrations, and,
therefore, some crucial components are subject to fatigue loads. One cannot really avoid fatigue
loads; they are as natural to machines as breath is to a human body. Therefore, we should learn
more about them so as to avoid surprise failures. It has been observed that most fracture failures
are initiated by fatigue loads. Initially, the crack length is subcritical and the crack is not dangerous.
With subsequent load cycles, the crack grows to acquire length close to the critical length and then
only conventional fracture mechanics (KIc , JIc , etc.) come into picture [9.1–9.3]. Mechanisms of
crack initiation, especially quantitative models, are still not known well. However, once a crack has
nucleated and has grown to the stage where it can be detected, some empirical or semi-empirical
formulations are available for its subsequent growth.
9.2.1 Terminology
There are two kinds of fatigue loads, (i) constant amplitude load and, (ii) variable amplitude load.
The loads on locomotive axles are of constant amplitude whereas fluctuating wind load on a wing
of an airplane is of variable amplitude. In certain cases, both types of loads may be superposed on
a component.
Figure 9.1 shows the constant amplitude loading with maximum stress smax and minimum stress
smin with the stress range Ds given by:
Ds = smax – smin
We define mean stress sm as:
s max + s min
2
Corresponding to smax and smin, we can determine Kmax and Kmin as
sm =
Kmax = f (a/W ) smax p a
(9.1)
Elements of Fracture Mechanics
Stress s
190
Ds
smax
smin
Cycle N
Fig. 9.1 Constant amplitude loading
Kmin = f (a/W ) smin p a
where f (a/W ) is the geometric factor for crack length a and component width W.
In most cases, dependence of f (a/W ) on crack length a is secondary in comparison to a
dependence. In order to keep the calculations simple, some designers prefer not to consider
variation of f (a/W ). The difference of Kmax and Kmin is an important parameter for determining
crack growth and is expressed as:
D K = Kmax – Kmin
Another parameter, stress ratio R, is also used and defined as:
R=
s min Kmin
=
s max Kmax
There are three categories of R — positive, zero and negative, as shown in Fig. 9.2. Positive R is
tension-tension fatigue whereas negative R is tension-compression fatigue. For negative stress
ratio, compressive stress loading is not likely to grow the crack and, therefore, some investigators
treat this case same as the one having R = 0. However, in sophisticated analysis, R = 0 and R < 0
may be treated differently; some introductory discussion is included in Sec. 9.7.
A fatigue crack may be initiated at an existing notch, an inclusion or a surface. It has been
observed that initiation requires a large number of load cycles. Once a crack is initiated, it grows by
some distance in every cycle, initially with extremely small growth per cycle. As the crack becomes
longer, the rate of propagation per load cycle, da/dN, also increases. Obviously, cracks of very
small lengths cannot be detected by available non-destructive test-techniques. The number of
cycles required to initiate a crack and then make it grow to a detectable length is known as initiation
life Ni. Detectable crack in most cases is still sub-critical and needs to grow further under the fatigue
load. The number of cycles required to grow the smallest detectable crack to a critical size is known
as propagation life Np. Thus, the total life N becomes
N = Ni + N p
Fatigue Failure and Environment-assisted Fracture
191
K
R>0
0
N
K
R=0
0
N
K
R = –1
0
N
Fig. 9.2 Three categories of stress ratio
9.2.2 S-N Curve
In the nineteenth century, many structural bridges collapsed and the axles of locomotives failed
owing to fatigue loads. It was then realized that a structure could fail even if the applied stress was
much smaller than the yield stress. Fatigue test-machines were then developed. A test-machine
applies a constant amplitude loading on a rotating shaft with bending stress (R = –1). Using the
machine an empirical relation is determined between applied stress (peak value of the fluctuating
load) and number of cycles N required to cause the failure. The relation is generally known as S-N
curve shown in Fig. 9.3 (a) for steel and Fig. 9.3 (b) for nonferrous metals. At a higher stress, of course,
the component has a shorter fatigue life. For steel, it is found that below the endurance limit se the
material does not fail. However, distinct endurance limit is not observed for nonferrous metals. It has
been a common practice to design a steel component such that the critical stress does not exceed se.
Now, more sophisticated models are being developed based on the concepts of fracture mechanics.
For nonferrous metals, conventional design approach has been to determine allowable stress sa for a
reasonable number of cycles, say 108 cycles, as shown in S-N curve of Fig. 9.3 (b).
S-N curves have been in use for more than a century, and are still being used by conventional
designers. S-N curves have certain limitations. An S-N curve adopts a black-box approach and it
does not explore the mechanisms of failure. It does not even distinguish between initiation life and
propagation life; only overall fatigue life is taken into account. There is hardly any consideration of
the specimen size; that is, data generated on small size specimen may not be applicable on large
size components. Also, the data on an S-N curve has a large scatter suggesting that the formulation
Elements of Fracture Mechanics
300
800
200
s, MPa
1200
s, MPa
192
400
se
100
0
102
104
106
108
sa
0
101
104
106
N
N
(a)
(b)
108
Fig. 9.3 (a) S-N curve for steel with endurance limit se and (b) S-N curve for non-ferrous
metals with no clear cut endurance limit
needs to be more rigorous. A component, designed on the basis of endurance limit, may still fail
during its use. Locomotive wheel-axles are checked frequently for fatigue crack growth because
S-N curve does not impart sufficient confidence for a fool proof performance.
The field of damage tolerance is being developed these days. Certain defects in a component can
be tolerated with a reasonable assurance on quality. Number of defects may increase or existing
defects may become longer with time. A non-destructive test is conducted to assess if the component
is still safe for further usage. However, designs based on S-N curve approach do not entertain damage
tolerance.
9.2.3 Crack Initiation
Crack initiation is observed to occur at the tip of an existing crack or at some point of a free surface.
A fatigue crack grows with each applied load cycle and therefore, crack growth per unit cycle,
da/dN, is an important parameter. Initially, da/dN is extremely small, may be as small as 10–10
m/cycle which is of the order of one lattice parameter. Such a small crack growth rate cannot be
detected easily and, therefore, it is included in crack initiation.
As the crack grows DK increases and da/dN eventually becomes quite large. When a fatigue
crack is about to become critical, da/dN may become as large as several millimeters per load cycle.
The crack growth is divided into three regions as shown in Fig. 9.4 on log-log scale. The crack
growth curve is known as Sigmoidal Curve. In the region I, da/dN is very small. In fact, there is no
initiation of crack growth if DK is smaller than a threshold DKth. Even for DK exceeding DKth, crack
Fatigue Failure and Environment-assisted Fracture
Region II
Region III
da/dN, log scale
Region I
193
DKth
DK, log scale
Fig. 9.4 Characteristic growth curve of a fatigue crack
TABLE 9.1
Representative value of DKth and its dependence on stress ratio R
Material
Mild Steel
Austenitic Steel
Maraging Steel
suts(MPa)
R-ratio
DKth(MPa m )
430
–1.00
0.50
0.75
–1.00
0.67
6.4
4.3
3.8
6.0
2.7
685
2010
growth per cycle cannot be easily detected. Threshold DKth depends on material properties and
stress ratio R; Table 9.1 shows representative values for some commonly used materials [9.4].
Initiation at the Tip of an Existing Defect
In case of a structural component, a crack may initiate at the tip of an existing defect, a reentrant
corner, a slot, a void or an inclusion. In these cases, there exists some DK that helps in the nucleation
of the fatigue crack. The initiation is influenced significantly by the radius of curvature of the
defect. An inclusion with sharp notch is likely to nucleate the fatigue crack easily. Some
investigators [9.4] have tried to investigate the effect of the radius of curvature on the initiation life
Ni. Figure 9.5 shows the variation for initiation life with DKI/ r where r is the radius of curvature
of the crack tip [9.5]. Clearly, Ni is high for blunt crack tips.
194
Elements of Fracture Mechanics
4000
r = 0.2 mm
0.4
DKI r, MPa
2000
1000
800
600
400
1.6
6
200
103
2
4
6 8 104
2
4
6 8 105
2
6 8 106
4
N
Fig. 9.5 Dependence of nucleation life on radius of curvature of the defect and
DKI for HY-130 steel
Initiation on a Surface
It is commonly observed that a fatigue crack initiates on a surface where no crack was existing
before the cyclic load was applied. In fact, fatigue load, being a powerful force, tries to first initiate
a crack on the surface of an existing inclusion, a hole or a reentrant corner. If any such initiation is
not favorable, the fatigue load initiates a crack on a free surface. What causes the initiation?
When a metal specimen is subjected to a cyclic load, it is observed that dislocation glide bands
initiate on the specimen surface. With increasing load cycles, these glide bands multiply laterally
and become denser. As a result, the surface no longer remains smooth. The dislocations transport
material from or to the surface, creating extrusions and intrusions on the surface as shown in Fig. 9.6.
An intrusion acts like a crack with a finite DK. If the initial surface is not very smooth it encourages
nucleation and growth of these intruded cracks. In fact, the surface of critical load bearing shafts is
polished to have a very low roughness. It pays to make the surface very smooth as it delays the
nucleation of surface cracks and enhances Ni. It is difficult to predict initiation life from the free
surface of a component because the micro-mechanism of crack nucleation is still not well
understood and quantitative formulations are difficult.
Smooth surface
Intrus
ion
Extru
sion
Fig. 9.6 Extrusion and intrusion created by glide bands
Fatigue Failure and Environment-assisted Fracture
195
9.2.4 Crack Propagation
The crack propagation rate, da/dN, depends on DK and stress ratio R and it can be expressed in as
da
= f (DK, R)
dN
Many attempts have been made to find the nature of this relation. The results are empirical or
semiempirical. For many materials, dependence of da/dN on R is not very significant, especially in
Region II as shown in Fig. 9.7. Most designers are contented with the simple form:
da
= f (DK )
dN
da/dN, log scale
R=0
0.4
0.8
DK, log scale
Fig. 9.7 Effect of stress ratio on Sigmoidal curve
Paris law is most widely used and is stated as:
da
= C (DK ) m
(9.2)
dN
where C and m are material constants to be evaluated through experiments for a material under
consideration. To have a feel of the material constants, representative relations for some materials
are:
∑ Ferrite-Pearlite steel [9.3]
da
= 6.8 ¥ 10–12 (DK ) 3.0
dN
∑ Martensitic steel [9.3]
da
= 1.33 ¥ 10–10 (DK ) 2.25
dN
(9.3)
(9.4)
196
Elements of Fracture Mechanics
∑ Austenitic stainless steel [9.3]
da
= 5.5 ¥ 10–12 (DK ) 3.25
dN
(9.5)
∑ Cast iron [9.6]
da
= 4.3 ¥ 10–8 (DK ) 4
dN
∑ Aluminum alloy (7075-T6) [9.6]
(9.6)
da
= 1.1 ¥ 10–11 (DK ) 3.89
dN
(9.7)
In these relations, crack length a is in meter and DK in MPa m . It is worthwhile to discuss the Paris
law further and obtain an expression for crack’s propagation life Np. Substituting DK into the Paris
law, we obtain:
da
= C [ f (a/W )Ds ]m (p a)m/2
dN
Rearranging and integrating, we have:
af
Ú
a0
a
-
m
2 da
f m ( a /W )
Np
= C( Ds )m p m /2
Ú dN
(9.8)
0
where a0 is the initial crack length of the propagating stage and af is final crack length. It is worth
noting that af need not be the critical crack length. Rather, it is taken usually to be the longest crack
length permissible in the damage tolerance design. If f (a/W ) does not vary strongly, it can be taken
as constant and then the expression can easily be integrated to give:
Np =
(- m2 +1) - a(- m2 +1)
a0
f
m
m
(m /2 - 1) C f ( a0 /W )( Ds )m p 2
(9.9)
Note that this expression is not valid for m = 2 and one should go back to Eq. (9.8), which is
simplified to:
af
Ú
a0
a-1 da
= C( Ds ) 2 p
f 2 ( a /W )
Np
Ú dN
0
On integration,
Np =
Ê af ˆ
ln Á ˜
Ë a0 ¯
(9.10)
C f 2 ( a0 /W )( Ds )2 p
Example 9.1 An edge crack, detected on a large plate, is of length 3.1 mm under a constant
amplitude cyclic load having smax = 310 MPa and smin = 172 MPa. If the plate is made of a ferritepearlite steel and KIc = 165 MPa m , determine (a) propagation life up to failure and (b) propagation
life if the crack length a is not allowed to exceed 25 mm.
Fatigue Failure and Environment-assisted Fracture
197
Solution: For this large plate with the edge crack
f (a/W ) = 1.12
Ds = 138 MPa and a0 = 0.0031 m
(a) The critical length ac = af = K 2Ic/[p (1.12 smax)2] = 0.0719 m
Taking value of C and m from Eq. (9.3) we obtain propagation life up to fracture using
Eq. (9.9) as
Np =
=
(0.0031)(
) - (0.0719)( - 32 +1) È
- 32 +1
(
3
2
)
-1
1
˘
Í 6.8 ¥ 10 -12 ¥ (1.12)3 (138)3 ¥ p 3/2 ˙
Î
˚
17.96 - 3.729
¥ 7153 = 203.6 ¥ 103 cycles
0.5
(b) Replacing af = 0.0719 by af = 0.025 in the solution of Part (a), we have
=
17.96 - 6.325
¥ 7153 = 166.4 ¥ 103 cycles
0.5
It is worth noting that difference in propagation life of two cases is small, only 37200 cycles. This
is so because the crack propagates faster at a higher crack length.
The S-N curve approach has been used for a very long time and to a certain extent it works. It is
worth exploring whether S-N curve is a subset of the Paris law. Focusing again on Eq. (9.9), we may
argue that af is governed by KIc and if the material is not changed from one experiment to another,
it remains constant. Similarly, a0 is the initial crack length which can again be taken constant if
specimen geometry is not altered during experimentation. We have then,
Np =
leading to
constant
( Ds )m
log Np = – m log Ds + constant.
This is a familiar linear S-N curve on log-log plot.
If the effect of stress ratio R is to be included, the Paris law has been modified by several
investigators. One such law, proposed by Forman et al. [9.7], is stated as
C( DK)m
da
=
dN (1 - R)Kc - DK
where Kc is the critical stress intensity factor. If Kmin is negative, the crack tip closes for compressive
portion of a stress cycle and therefore Kmin is taken as zero (R = 0).
The complete Sigmoidal curve of crack propagation can also be expressed through one equation.
Erdogan and Ratwani [9.8] have suggested the following expression:
198
Elements of Fracture Mechanics
m
n
da C(1 + b ) (D K - D Kth )
=
dN
Kc - (1 + b ) D K
where c, m, n, DKth and Kc are material constants, and
b=
Kmax + Kmin
Kmax – Kmin
Some investigators argue that the plastic zone size plays an important role in actual crack
propagation, and in such a case elastic-plastic fracture mechanics should be used. One such form
of crack propagation law is
da
= C (D J )m
dN
where J is J-Integral. A similar analysis may be carried out with CTOD parameter or effective crack
length.
9.2.5 Effect of an Overload
An overload is a pulse (or a set of pulses) of higher amplitude on a constant amplitude fatigue load as
shown in Fig. 9.8. The crack propagation rate retards considerably after the overload pulse [9.9, 9.10].
s
Overload
pulse
Retarded
growth
N
Fig. 9.8 An overload pulse on the constant amplitude fatigue load
Why does the propagation rate decrease after the overload pulse? To understand it, we look into
the behavior of the plastic zone near the crack tip. The overload forms a much larger plastic zone.
During the unloading portion of the overload pulse, the surrounding elastic material tries to regain
its original state. However, the plastic zone cannot regain the original state and, therefore,
compressive residual stresses are developed in the vicinity of the crack tip. In the follow-up pulses
of the constant amplitude load, the tensile loading at the crack tip is suppressed by the residual
compressive stresses. Consequently, da/dN is lowered appreciably. The crack tip may remain
within the enclave of residual compressive stress for many subsequent pulses of the constant
amplitude load as shown in Fig. 9.9.
Fatigue Failure and Environment-assisted Fracture
199
Enclave of residual
compressive pulse
Plastic zone
of a subsequent
pulse
Overload
pulse
Fig. 9.9 Enclave of residual compressive stress developed by the overload and
plastic zone of subsequent pulses
Crack length, a
It, therefore, takes some load cycles for the crack tip to come out of the enclave and then only
da/dN regains its value based on the length of the crack and DK at that location. The effect of
overload on crack growth is shown in Fig. 9.10.
Growth
without
overload
Retarded
growth
Overload
pulse
N
Fig. 9.10 Retarded growth of the crack owing to the overload pulse
9.2.6 Crack Closure
We will first consider the cases when Kmin is zero (R = 0). During the unloading portion of a load
cycle, it has been observed that crack faces start touching each other for some finite tensile stress.
Figure 9.11 shows that the crack closes at a point A. When the stress is reduced further to zero,
Elements of Fracture Mechanics
200
compressive stress at the crack tip is developed [9.11]. Why does a crack close at point A and not at
point B? To understand it, we should study the nature of cracked surfaces developed through the
fatigue load. Each cycle produces some crack extension forming a ridge and a valley, known as
striation. Furthermore, surfaces of a striation are not smooth. They are rough owing to non-uniform
local plastic deformation during separation. It is quite improbable to expect that all the peaks of the
top surface will fit exactly into the valleys of the bottom surface on unloading. Therefore, as K is
reduced below point A of Fig. 9.11, the touched surfaces start exerting compressive load. The early
crack closure is facilitated if the newly created surfaces are oxidized or have some other chemical
reaction with the atmospheric air. Moreover, in practical situations it is rare to have only pure
Mode I acting on a crack. Some contribution of Mode II, Mode III is generally present that causes
lateral shift of cracked surfaces and would not allow perfect matching of the surfaces on unloading.
Crack
closure
DKeff
K
A
Kcl
B
N
Fig. 9.11 Early closure of a crack and reduction in DK to DKeff
Fortunately, the crack closure is helpful in retarding the crack growth. It is clear from Fig. 9.11
that effective DK is reduced due to early closure. Therefore the crack propagation rate is reduced
and a component has a longer fatigue life. Early crack closure also works for R < 1 and may work
for positive R as shown in Fig. 9.12. Clearly for high positive R crack closure is less.
K
K
DKeff
DKeff
A
A
Kcl
Kcl
0
N
(a)
0
N
(b)
Fig. 9.12 (a) Crack closure for R > 0, and (b) for R < 0
How to find the effective DK? One can perform experiments to find K at which the crack closes.
However, it is not possible to do experiments every time a component is designed. We would
rather have some models to determine it. The effective DK is expressed as
(DK)eff = Kmax – Kcl
Fatigue Failure and Environment-assisted Fracture
201
where Kcl is the stress intensity factor at the time of closure of crack faces [9.12, 9.13]. However, Kcl
is generally not determined. An alternative procedure is adopted by modifying the Paris law by
associating a factor U with DK. The Paris law is then written as:
da
= C (UDK)m
dN
(9.11)
where
U=
Kmax - Kcl
Kmax - Kmin
Several empirical models have been suggested for finding U. Elber [9.12] proposed the formula for
2024-T3 aluminum as
U = 0.5 + 0.4 R
and Schijve [9.13] suggested the formula for 2024-T3 aluminum, as
(9.12)
U = 0.55 + 0.33 R + 0.12 R2
(9.13)
Crack closure has already been accepted by industry and some computer packages do make
necessary correction on crack propagation rate.
9.2.7 Variable Amplitude Fatigue Load
Constant amplitude fatigue load is simpler to study but is not as common as variable amplitude
load in many applications. Machines and structures like pumps, vehicles, earth moving
equipments, rolling mills, ball and cylindrical bearings, airplanes, bridges, ships and off-shore
structures are subject to variable amplitude fatigue loads.
Can the Paris law or some other law for crack propagation rate still be used? Yes, but with some
reservations. The data of the variable amplitude load should be processed before invoking the
Paris law [9.3, 9.8]. However, there exists a problem of fundamental nature. We have already noted
in Sec. 9.6 that a high stress pulse retards the propagation rate of some subsequent load cycles
significantly. The number of retarded load cycles depends upon the relative magnitude of stresses.
In case of variable amplitude loads there are many overload pulses, each having associated load
cycles with retarded crack growth. It is difficult to keep track of the extent of retardation and
predict the crack growth. However, if the variable amplitude is limited to a narrow band, the
retardation of the crack growth may be ignored. This simplification is on the conservative side and
a designer may not mind it.
To predict the life or the crack growth of a component, one should know the expected load history
based on previous experiences. It is worth noting that the nature of load varies considerably from one
kind of application to another [9.14]. For example, variable amplitude load on aircraft wing is quite
different from the load on a lifting crane. Moreover, the load fluctuations generally do not follow the
Gaussian distribution. Based on the application, a suitable statistical procedure is adopted to
determine root mean square value of DK. If Paris law is chosen, the propagation equation becomes:
da
= C(DK rms )m
dN
202
Elements of Fracture Mechanics
Example 9.2 Fluctuating load on a critical component of an offshore structure is shown by a
histogram in Fig. 9.13. During a routine check-up, an edge crack of length 1.5 mm is detected. If the
crack length is not allowed to exceed 25 mm, determine the remaining life of the component. Use
Paris law with material constants as C = 6.0 ¥ 10–12 (MPa)–3.2 m–0.6 and m = 3.2.
% Occurrence
40
30
20
10
0
60
80
100
120
140
160
Ds, MPa
Fig. 9.13 Percentage occurrence of load pulses of the variable amplitude load
Solution: (Ds)rms is evaluated as
(Ds)rms =
0.2 ¥ 902 + 0.4 ¥ 1102 + 0.3 ¥ 130 2 + 0.1 ¥ 1502 = 117.4 MPa
da
= C(DK rms )m
dN
Using Eq. (9.9),
Np =
=
0.0015 (
-3.2
2
) - 0.025 ( -3.22 +1)
+1
( 3.22 - 1) ¥ 6.0 ¥ 10-12 ¥ (1.12)3.2 ¥ (117.4)3.2 ¥ p
3.2
2
49.47 - 9.146
= 297 ¥ 103 cycles
-6
135.6 ¥ 10
9.3 ENVIRONMENT-ASSISTED F RACTURE
9.3.1 Introduction
Man makes various machines, structures, and other mechanical systems, which deteriorate slowly in
their service environment. In fact, all kinds of atmosphere, most kinds of water, oils, chemicals, gases,
and food corrode structural components. The corrosion is known to be high in the vicinity of a crack
tip where stresses are large and the plastic deformation is intense. In metal components, slip bands
with innumerous dislocations facilitate the dissolution of material and thus the crack length increases
slowly. This process may also be facilitated by the anodic dissolution at the crack tip. For example,
when a component of mild steel is submerged in water, it loses its metallic ions close to the crack tip.
Fatigue Failure and Environment-assisted Fracture
203
The loss of metallic ions leaves extra electrons on the surface, thus making the crack tip behave as an
anode. The dissolution through such an electrolytic cell (galvanic cell) is known to be accelerated
under a stressed field. Also, owing to the tensile field at the crack tip, small sized hydrogen atoms
may diffuse into the component’s material, causing hydrogen embrittlement and easy growth of the
crack. Thus, the environment–assisted corrosion gradually increases the length of a subcritical crack.
The growth of a crack length, in turn, increases the Stress Intensity Factor (SIF) and a stage reaches
when the SIF becomes critical, causing rapid failure of the component. A designer should choose the
material of the component or the working environment carefully to limit the environment-assisted
crack growth.
It is difficult to formulate the environment-assisted fracture mechanics. It is an interdisciplinary
field involving chemistry of materials, electrical aspects of current flow, and various mechanical
and materials aspects such as diffusion, plastic deformation, grain boundaries, impurity content,
voids, etc. Many different kinds of corrosion mechanisms act on the tip of a crack. Some
mechanisms are known and backed by experimental findings. Models for some other mechanisms
are available in literature but they have not been verified still by experimental. The complexity of
the formulation increases further as there are many different kinds of materials and a variety of
corrosive environments. The corrosion of a material is often found to be affected considerably by a
slight change in a material parameter such as impurity content, percentages of alloying elements,
heat treatment, work-hardening, etc. Also a slight change in environment may influence the
corrosion substantially. For example, atmospheric humidity plays a crucial role in the rusting of
steel components. Consequently, a comprehensive coverage on environment-assisted cracking is
beyond the scope of this book and only an overview is presented in this chapter.
A corrosive environment affects all surfaces of a component. Even a smooth surface develops
roughness under the combined effect of high stresses and environment. The micro-crevices thus
generated grow with time. Some of them become cracks of finite length and grow further to cause
component’s failure. Similar to the fatigue failure from a smooth surface, there is an incubation time
for an environment-assisted crack to nucleate. In this chapter, owing to the complexity involved we
will not discuss nucleation of cracks on smooth surfaces. We would be confining our attention only to
an already existing crack in a metal component to study how a corrosion crack nucleates on the tip of
the existing crack and how it grows to cause the eventual failure of the component.
9.3.2 Micromechanisms
Most structural metals like aluminum, titanium, steel, copper, etc. are naturally protected from a
thin oxide layer on their external surfaces. The thin layer, being passive in most metals, protects the
interior material from rapid corrosion. For example, if an aluminum component is cut in air, a thin
layer of aluminum oxide is immediately developed on the new surfaces and protects the
component. For stress corrosive cracking, the passive layer should be broken. As discussed in the
earlier chapters, the stresses at the crack tip are high, which cause intensive plastic deformation in
the plastic zone. In metals, slip bands consisting of innumerous dislocations make the surface near
a crack tip rough with micro-cracks. Consequently, the passive thin layer is broken, exposing the
material to the corrosive atmosphere. In fact, the size of the plastic zone is not important because
only the plastic deformation near the crack tip is influenced by the corrosion. Thus, the rate of
crack growth does not depend much on increasing stress intensity factor with the advancement of
the crack length. This aspect will be discussed further in the next section.
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Elements of Fracture Mechanics
The crack surface in the vicinity of a crack-tip, as stated in the previous section, becomes the anode
of the electrolytic cell formed. If the crack surface of steel component is wet with water, the iron ions
in the vicinity of the crack tip dissolve, leaving excess electron on the surface. Thus the crack surface
near the tip starts acting as the anode of the electrolytic cell as modeled in Fig. 9.14. Far away from the
crack tip, the passive layer works as the cathode of the cell. The excess electrons on the anode pass
through the conducting body of the component to the cathode surfaces. These electrons then reduce
the hydrogen ions of the surrounding water, resulting into the evolution of hydrogen gas. Some of
these hydrogen atoms may diffuse into the component to cause hydrogen embrittlement which is
discussed subsequently in this section. The current density of the electrolytic cell regulates the rate of
material loss at the crack-tip. Since the metal is in direct contact with the corrosive environment near
the tip, the current density can be high. However, the resistance of the cathode oxide layer is quite
high and thus it restricts the current density of the electrolytic cell and the dissolution rate to low
levels. It is worth noting that the crack growth through anodic dissolution is continuous.
Fig. 9.14 The electrolytic cell formed near the crack tip
It is also observed that in comparison to the interior crystalline material the grain boundaries of
most metals are more anodic. This causes higher dissolution rate at the grain boundaries, resulting
into intergranular crack growth.
Owing to the tensile stresses in the vicinity of the crack tip of a component, the hydrogen atoms,
being small in size, diffuse into the grain boundaries, the voids, the inclusions and highly strained
slip-bands. The diffused hydrogen atoms make the material brittle and the process is known as
hydrogen embrittlement. The diffusion is usually slow but makes the material in the vicinity of the
crack tip brittle and the crack is no longer able to withstand the SIF. Consequently, the crack grows
to the point where the effect of hydrogen embrittlement is diminished. The crack thus moves by a
small step. Usually the failure is intergranular as grain boundaries are more susceptible to
hydrogen embrittlement. It again takes some time for hydrogen atom to diffuse beyond the
extended crack tip. Thus, unlike the continuous anodic dissolution the crack, the growth through
hydrogen embrittlement is in short steps.
In some cases, the anodic dissolution facilitates the diffusion of hydrogen. For example, anodic
dissolution grows a crack tip within a grain till it reaches a grain boundary, thus exposing the grain
boundary for hydrogen diffusion. Thus the environment-assisted crack growth is often caused by
the simultaneous actions of several mechanisms.
Fatigue Failure and Environment-assisted Fracture
205
9.3.3 Test Methods
Tests on environment-assisted cracking fall into two categories: (i) the constant load method and
(ii) the constant displacement method. In both categories, specimens with a precrack are employed
whose notch is sharpened with a fatigue growth. The constant load test setup requires multiple
specimens and thus the testing expenses are on higher side. On the other hand, only one specimen
can suffice in the constant displacement test, but the crack does not grow to become critical to
cause the failure and therefore the experiment yields less number of results. Details of strengths
and limitations of each method will be discussed subsequently in this chapter.
Constant Load Method
In the constant load test, a cantilever specimen with a sharp notch is commonly used as shown in
Fig. 9.15. A lever arm is bolted to the far end of the specimen so as to generate high bending moment
using a dead load P. The bending moment remains constant throughout the duration of an
experiment. But the Stress Intensity Factor (SIF) increases with the advancing crack. The crack tip
is surrounded within a corrosion chamber where corrosive environment is controlled as per the
requirements.
Fig. 9.15 A schematic test setup of the constant load method
The crack tip is prepared following the same procedure as used for making specimens for
KIc-test (Sec. 8.2.4). A machine crack is prepared, preferably with a Chevron notch to initiate the
fatigue crack at small number of cycles. The fatigue load is carefully chosen to have the SIF at the
crack tip substantially smaller than the SIF developed by the constant load during environmentassisted testing.
A standard test machine such as a universal testing machine may be employed but corrosion
test are long duration tests and therefore use of a costly machine may be avoided. A simple setup as
shown in Fig. 9.15 can be easily designed and developed in a test laboratory. The constant load test
yields several results such as:
∑ Determination of time-to-failure for an initial SIF, KIi
∑ Determination of threshold SIF, KIth, below which the precrack would not grow
∑ Determination of the rate of crack growth, da/dt.
It is worth noting here that most real life structures of aerospace, automobile, nuclear plants, etc.
are not thick enough to qualify for plane strain conditions. Also the results are required for a
206
Elements of Fracture Mechanics
specific set of material-environment interface. Since there are many combinations of materials and
environments, it is practically very difficult to have data for all cases in the open literature. Thus a
user is expected to determine results for the combination of material and environment to which the
component would be subjected in service conditions. In such a circumstance, the user may get the
testing done under the conditions which simulate the actual application of the component. Thus,
he need not make specimens which satisfy stringent conditions of plane strain. Thinner specimens
corresponding to plane stress or transitional cases (Fig. 5.8) may be used which will fail at the
maximum value of SIF, Kmax. Of course, in case a specimen meets the requirements of plane strain,
the specimen will fail at KIc.
A set of multiple specimens, usually 5 to 10, are employed to conduct the constant load test. The
specimens are loaded with varying initial stress intensity factor, KIi. During the experimentation
the crack length of a specimen is continuously monitored. To determine the length of a crack it is
not recommended to use the usual method of monitoring the change of electrical resistance with
the increasing crack length. The resistance method applies a voltage on the specimen which may
affect the voltage of the electrolytic cell formed at the crack tip. Thus the resistance method is likely
to introduce errors in experimentation. One usually adopts the method of monitoring the change
in compliance by recording the crack mouth opening displacement and then calculating the crack
length.
It usually takes a fairly long time for the precrack to start growing once a specimen is placed in a
corrosive environment. This time is known as incubation time, tinc. After the incubation time, it takes
an additional time tg for the crack to grow till failure. Thus the total time-to-failure tf becomes:
tf = tinc + tg
The characteristic relation between time log(t) and KIi is shown in Fig. 9.16. If such a relation is
available to a designer and the initial crack length of a crack is known, he can estimate the life of a
component.
log (t)
tg
tinc
KIth
KIi
Kmax
Fig. 9.16 Characteristic relation of incubation time tinc and
growth time tg versus initial SIF, KIi
Fatigue Failure and Environment-assisted Fracture
207
A relation between KIi and time-to-failure tf is also explored through constant load tests to
determine threshold KIth , below which no crack extension takes place due to the stress-assisted
corrosion. Figure 9.17 shows a characteristic relation between KIi and time to failure. It is, in fact,
difficult to determine KIth as tests have to be performed for very long durations. Some experiments
are conducted by applying a load which develops SIF just above the KIth. Then load is reduced to
have no crack growth even when the specimen is exposed for a long time. What should be the
adequate test duration to determine KIth? For some material-environment combinations, 1000 hours
of test time is adequate to assume that the crack would not grow if the specimen is exposed for longer
time. But for other combinations it may require testing for longer duration. In fact, the adequate
duration of a test depends on the material-environment interface. Therefore for testing an altogether
new combination, preliminary testing is required to determine the adequate test duration.
KIc
Kmax
KIth
log (tf)
Fig. 9.17 Characteristic relation between initial SIF and time-to-failure, tf
The constant load tests also leads to another useful relation between the rate of crack growth and
the instantaneous SIF that increases with the crack growth. Figure 9.18 shows the schematic diagram
of log(da/dt) versus KI where a is the crack length. In most combinations of materials and
environments, three regions are observed. An initial crack having its SIF less than KIth does not grow
at all. In Region I, there is a steep rise in da/dt with KI. In the Region II, da/dt shows a plateau; that is,
there is no increase in the crack growth rate with the increasing SIF. As mentioned in the previous
section, the corrosion depends only on the material in the vicinity of the crack tip. At a higher SIF the
plastic zone size increases, but the flow stress at the crack tip is not affected much. Thus stress
corrosion cracking in the Region II continues to grow at the constant rate till the SIF starts approaching
Kmax. The rate of crack growth increases at a much faster rate in the Region III and causes the failure
of the specimen. The rate of increase in da/dt is so high in certain materials that the Region III is not
even recorded. However, the Region III is not of much use to us and we need not make special
arrangement to monitor it. We thus note that in the Region I a stress induced corrosion crack is
nucleated at the tip of the existing precrack. The Region II signifies the growth period of the crack.
The Region III is not important as the crack propagates rapidly to cause the failure of the component.
The crack growth rate in the Region II is usually in the range of 10–8 mm/s to 10–4 mm/s for most
material-environment interfaces. It is worth noting that da/dt ª 10–8 mm/s corresponds to about 0.3
mm crack growth per year which is negligible for most applications. There is no need to do analysis
208
Elements of Fracture Mechanics
Region II
III
log (da/dt)
I
KIth
KI
Fig. 9.18 Characteristic relation between the rate of crack growth and SIF
on such material-environment interfaces. On the other hand da/dt ª 10–4 mm/s corresponds to 9
mm growth in a day which is quite high and the designer should do some major design changes
such as exploring other material-environment combination. For intermediate cases of da/dt in the
range of 10–6 and 10–7 mm/s, a designer may choose his design parameters to either avoid failure
within the designed life of a component or recommend frequent checking of the crucial
components through appropriate nondestructive tests.
Constant Displacement Method
In the constant displacement method, usually a compact test specimen is modified. An initial
displacement is given between the two ends of the cantilevers of the specimen and then the
displacement is maintained (fixed grip loading). Initially the SIF is high but it decreases as the
environment-assisted crack grows. A universal test machine can be used to conduct this test but
usually a simple configuration is incorporated within the specimen. Figure 9.19 shows the schematic
diagram of the configuration in which a tapped hole is made in the top cantilever. In the bottom
Bolt
Pre-crack
Displacement gauge
Load distributor
Fig. 9.19 A schematic test setup of the constant displacement method
Fatigue Failure and Environment-assisted Fracture
209
cantilever a through-the-thickness hole is made within which a rod with its top flattened is slipped
in. In fact, this hole is made before the machine crack is introduced in the specimen so as to facilitate
its fabrication. This rod distributes the point load applied by the bolt that is screwed into the tapped
hole in the top cantilever. By rotating the bolt the bottom cantilever is pushed out to develop the
required initial gap between the two ends of the cantilevers. The specimen is placed within an
environmental chamber or it can be taken to an outdoor field to simulate the conditions of the actual
application.
The crack length is monitored to determine the crack growth rate. The length at which the crack
stops growing corresponding to KIth and thus one single experiment yields KIth unambiguously.
Recall that in case of the constant load method, several tests are required at KIi slightly above and
slightly below the KIth. Even after running an experiment for a long duration, there is still a doubt that
the crack would have grown if the experiment had been conducted for a longer duration. However,
there is a major difference between the two kinds of experiments. In the constant load method the SIF
increases till failure. On the other hand, the SIF decreases in the constant displacement test to KIth and
thus the test does not evaluate the time-to-failure and the incubation time.
9.3.4 Major Factors Influencing Environment-assisted Fracture
There are many factors which influence the crack growth in a component under the combined
action of the stresses and environment. Some of the important parameters are:
Alloy chemistry: The environment-assisted cracking in materials often depends on the percentage
of the alloying elements. In certain materials even minor changes in the alloying elements
drastically alters the environment-assisted cracking. For example, the rate of crack extension in
aluminum 7079-T651 is about 1000 times greater than that of Aluminum 7178-T651 when tested in
3.5 % sodium chloride water [9.15].
Heat treatment: The yield stress of many alloys is enhanced by proper heat treatments, but the
increase in the yield stress may deteriorate the corrosion resistant. For example, in a study the alloy
steel 4340 was tested under the flowing sea water conditions [9.15]. When the heat treatment was
changed to increase its yield stress from 900 to 1400 MPa, decrease in KIc was not much, 95 to 66
MPa m only. However, the threshold SIF, KIth , dropped drastically from 76 to 10 MPa m . In
another study [9.15] on over-aging of the aluminum alloy 7079-T651, it was found that the rate of
crack growth decreased by a factor of 100 when the over-aging was increased from 1 hour to 15 hours
at 160° C.
Humidity: The humidity of environment has pronounced effect on environment-assisted fracture of
certain materials. In the aluminum alloy 7075-T651, it was found that the crack growth rate increased
24 times when the relative humidity of the surrounding air was increased from 5% to 67% [9.15].
Salt concentration: Service components are often immersed in a liquid with a salt whose
concentration may have marked influence on the corrosion behaviour. For example, an increase in
the concentration of potassium iodide enhances the crack growth rate (da/dt) in aluminum alloy
7079-T651 as shown in Table 9.2 [9.16].
210
Elements of Fracture Mechanics
TABLE 9.2 Variation of crack growth rate with the concentration of potassium iodide
Concentration of KI
(Mole)
da/dt
(mm/s)
0.002
0.020
0.200
2.000
3 ¥ 10–5
7 ¥ 10–5
80 ¥ 10–5
600 ¥ 10–5
Temperature: For some combinations of materials and environments, there is a considerable
influence of the temperature on the stress assisted cracking. If the influence is considerable, a
relation may be explored to do accelerated tests at higher temperatures and predict the behavior
on a longer time span at lower temperatures of components in service.
Work-hardening: Plastic deformation in a metal generates a large number of dislocations which in
turn enhance the corrosion. It is a common observation that a steel wire corrodes first at bends
where material is work-hardened. Some designers like to choose cold rolled sections and sheets as
materials of their components because their yield stress is significantly higher over that of hotrolled sections, about 20 % in mild steels. However, higher susceptibility to environment-corrosion
of work-hardened materials offsets the advantage of the higher yield stress.
9.3.5 Liquid Metal Embrittlement
Liquid Metal Embrittlement (LME) is another mechanism of environment-assisted crack growth.
On the surface of a metal component, a thin layer of some other liquid metal may cause the rapid
growth of a crack. In some cases the growth rate may be as high as 5000 mm/s. The exact
mechanism is not known but some investigators believe that the layer of a liquid metal around the
crack tip reduces the cohesive strength of the molecules of a ductile material, causing rapid growth
of the crack. Not all combinations of metals exhibit liquid metal embrittlement. Table 9.3 shows
which combinations of liquid metals and common structural materials are sensitive to LME.
TABLE 9.3 LME of a liquid metal layer on structural materials
Aluminum
Copper
Iron
Titanium
Mercury
Zinc
Lead
Tin
yes
yes
yes
yes
yes
no
yes
no
no
yes
yes
no
yes
yes
no
no
9.3.6 Design Considerations
In the design of structural and machine components, designers often do not give serious
consideration to fracture failure due to combined effect of stress and environment. In many cases,
Fatigue Failure and Environment-assisted Fracture
211
environment-assisted crack growth may be so slow that the crack may not become critical during
the life span of the component and thus need not be considered. It is worth noting here that not all
kinds of failures (strength, buckling, fatigue, creep, deflection beyond a certain stage, wear, etc.)
are analyzed in detail during the design of a component and only the relevant parameters are
identified and pursued. There are significant numbers of real life situations in which environmentassisted cracking may cause the failure of components in service conditions. These cases should be
analyzed carefully for stress-assisted corrosion.
Threshold stress intensity factor KIth seems to be attractive property for safe design. However
for most combinations of environment and material, KIth is small, of the order of 5–15 MPa m . If
a component is designed to limit the SIF of its cracks to KIth, the design will be very conservative
and thus expensive and impractical. For practical and economical designs, crucial components
should be checked at a regular interval through nondestructive techniques (Chapter 12) to identify
cracks which may become critical and cause the failure. In the modern industrial world, long
pipelines carrying various liquids or gases, railway tracks, axle of railway coaches and wagons,
piping of chemicals plants, etc. are routinely checked during the service conditions to identify
potentially dangerous cracks, which might have developed during the usage. Since these cracks
may grow and fail the components, necessary remedial actions are taken.
9.4 ENVIRONMENT-ASSISTED F ATIGUE FAILURE
The formulation of fatigue failure in an inert environment is still not well developed as well as the
environment-assisted growth of a stationary crack is also not formulated well due to many
parameters involved. The combination of the two makes the formulation still more complex and
causes the failure of a service component earlier than expected. Furthermore, often a synergy effect
exists between fatigue crack growth and environment-assisted cracking. For example, the
formation of a passive layer on the newly created surfaces of an advancing crack may get affected
by the fatigue load. In fact, the fluctuating load may break the new passive layer or delay its
formulation, resulting into higher crack growth.
A designer faces difficulties in predicting the life of a component loaded with combined effect
of stress, cyclic load and corrosive environment as reliable models are not available to him. He
may get tests conducted in simulated service conditions of the component. The tests are not only
expensive but takes very long time, weeks, months or years depending on the application. The
accelerated tests are still in the preliminary stages as it is difficult to correlate their results with
actual conditions of outdoors usages. The fatigue failure of most commonly used metals like
steel, aluminum and titanium is even enhanced considerably by the presence of water molecules
in air. And it is not practical, for most cases, to limit the humidity of surrounding air of structural
components. In fact, a certain minimum humidity is necessary for the comfort of people and thus
it is not justified to create the environment of dry air around a machine or a structure. Figure 9.20
shows the dependence of the crack extension per cycle in 7075-T6 aluminum alloy on the water
content in the environment [9.15]. There is a marked influence of water molecules on the crack
growth rate. When water content was increased from 2 ¥ 10 to 2 ¥ 104 ppm, the crack growth rate
increased from 2.6 ¥ 10–5 to 18 ¥ 10–5 mm/s—a substantial increase of seven times.
212
Elements of Fracture Mechanics
da/dN, mm/cycle
10– 4
10– 5
10– 6
103
102
10
104
105
Water content, ppm
Fig. 9.20 Dependence of the crack extension per cycle in the aluminum alloy 7075-T6
on the water content in the environment for Kmax = 13.75 MPa m
Many variables which have significant effect on environment-assisted cracking have already
been discussed in Sec. 9.3.4. However environment-assisted cyclic load adds two more variables to
the list of large number of variables, the frequency and the mean stress. The frequency is found to
have a considerable effect in many cases. Often a low frequency loading causes more
environmental corrosion, probably because it gives more time to the corrosive medium to work on
the crack tip. Experiments on 12Ni-5Cr-3Mo steel showed no dependence of crack growth rate per
cycle (da/dN) on frequency when the specimens were tested in air between slow rate of 0.1 and fast
rate of 10 cycles per second (Fig. 9.21). However when specimens were tested in the medium of 3%
NaCl at the high frequency load of 10 cps the relation between da/dN and DK was not too different
from the results of testing in air. As the frequency was reduced to 0.1 cps, there was a considerable
shift in the relation towards higher rate of crack growth [9.15].
10–3
0.1 cps
da/dN, mm/cycle
1
5
10
0.1–10 cps
2
Air
10–4
3% NaCl
20
40
DK, MPa
60
100
(m)1/2
Fig. 9.21 Variation in the crack extension per cycle with frequency in steel alloy
12Ni-5Cr-3Mo tested in 3% NaCl solution
Fatigue Failure and Environment-assisted Fracture
213
In short, the combined effect of a cyclic load and a corrosive environment hastens the crack
growth rate in many materials. A designer is recommended to look into the chances of
environment-assisted fatigue failure. Furthermore, one should be careful in generating the data on
fatigue failure. May be, the environment is also playing a significant role in the results as even the
atmospheric humidity may have a considerable effect on the crack growth. This raises a doubt that
probably early investigators, say before the year 1990, were not conscious of the contribution of
environment while generating data on fatigue crack growth and we may reevaluate some of the
important data.
9.5 CLOSURE
Fatigue loads and corrosive environment are two harsh service conditions on a structural
component. A subcritical crack, which would not grow under the controlled and ideal test
conditions of a laboratory, may grow under these harsh conditions. The crack length eventually
will increase to its critical length, followed by its rapid growth to the failure of the component. The
formulation of crack nucleation and growth is quite complex because many micro-parameters are
involved. Only a limited success has been achieved so far.
It takes a large number of cycles to nucleate a fatigue crack at the tip of an existing crack.
Similarly it takes a fairly long incubation time for an environment-assisted crack to appear at the
tip of an existing crack. In fact, satisfactory analyses have not been so far developed to predict the
nucleation of cracks under a fatigue load or in a corrosive environment or both. The irony is that
the time required to nucleate a crack is quite long and is comparable to the duration a nucleated
crack takes to propagate to the final failure of the component. Thus it is difficult to predict the total
time to fracture and therefore a crucial structural component should be checked through nondestructive techniques on a regular interval to detect nucleated cracks. The propagation time is
handy to give us some convenient breathing space between the two consecutive checks.
In case of a fatigue load, the crack proportion rate per cycle is determined through an empirical
law such as the Paris law. The propagation rate depends on DK and the stress ratio but experiments
show that DK plays a dominant role. An overload pulse develops compressive residual stresses in
the vicinity of the crack tip and therefore it retards the crack propagation rate significantly of some
load cycles following the overload pulse. Further, crack closure is another phenomenon that
reduces DK because the crack closes prior to complete unloading.
In environment-assisted fracture, several kinds of mechanisms act on a crack tip. In metals, an
electrolytic cell may develop with the crack tip acting as the anode. The metals near the crack tip
dissolves as metal ions leave the component to the electrolytic solution and thus the crack length
increases slowly. Another mechanism is the hydrogen embrittlement. The hydrogen atoms, being
small in size, may diffuse into the metal near the crack tip because the stresses are tensile in the
plastic zone. Consequently, the metal becomes brittle in the vicinity of the crack tip and the crack
grows easily. Unlike the fatigue crack growth rate which depends on the increasing DK with an
advancing crack, the environment-assisted crack growth is found not to depend on the increasing
SIF of a growing crack.
Environment-assisted fatigue deals with the combined action of fatigue loading and
environment corrosion, resulting into the faster growth of a crack. In some cases, there exists a
synergy effect between the fatigue loading and the environment-assisted cracking, increasing the
crack growth rate further.
214
Elements of Fracture Mechanics
QUESTIONS
1. Give three applications each of constant amplitude fatigue load and variable amplitude
fatigue load.
2. What are the shortcomings/limitations of S-N curve approach?
3. Why is it difficult to formulate crack initiation of a fatigue crack?
4. How does a fatigue crack get initiated on a smooth surface?
5. In case of negative stress ratio, smin is usually taken to be zero. Why so?
6. Propagation rate of a fatigue crack, in general, depends on DK and R but the Paris law,
which ignores the effect of R, is very popular. Why so?
7. Some investigators claim that DJ should be used in place of DK in a crack propagation law
for a fatigue crack. Is the claim justified?
8. How does an overload retard the growth of a fatigue crack?
9. What is crack closure? Why does it happen?
10. How is crack propagation rate determined for variable amplitude fluctuating load?
11. How do we account for retardation of a fatigue crack growth owing to overloads in variable
amplitude fatigue load?
12. Are predictions of crack growth accurate enough for engineering applications considering
the present know-how?
13. Why does the environment-assisted cracking occur mostly through inter-granular growth?
14. What is anodic dissolution?
15. Why does the rate of anodic dissolution depend on the resistance of the passive layer?
16. Why is it difficult to determine KIth through the constant load testing?
17. Why does work-hardening cause higher rate of crack extension of environment-assited
cracks?
18. Does the rate of crack growth of a corrosion crack not depend on the SIF for most materials?
19. It is difficult to design components which would not fail due to fatigue loads or stress
induced environment-cracking or both. Therefore a crucial component should be routinely
checked to identify whether cracks, which have potential to grow and cause the failure of
the component, have developed since the last check-up. Comment on these statements.
PROBLEMS
1. Determine nucleation life if a slot is made in a large plate (HY-130 steel) having a tipradius of 2 mm and a length of 40 mm from one edge to another. The plate is subjected to
a fatigue load of smax = 140 MPa and smin = 0.0 MPa.
2. An inclusion is detected in a plate of HY-130 steel through X-ray. It is modeled as a throughthe-thickness crack of length 22 mm, with the radius of curvature of the tip as 0.2 mm. If
the plate is subjected to a fatigue load of constant amplitude with s max = 140 MPa and
smin = 0.0 MPa, determine the number of cycles required to nucleate a crack at the tip of
the inclusion.
3. Determine the propagation life for the case of Problem 1 if the crack is not allowed to
exceed 60% of the critical length corresponding to KIc = 150 MPa m . Use the Paris law
with C = 7.2 ¥ 10–12 MPa–3 m–1/2 and m = 3.0. Also determine the total fatigue life (take slot
length as initial crack length).
Fatigue Failure and Environment-assisted Fracture
215
4. The Paris law of fatigue growth of a crack is known to have the form
da
= C (DKI) 3
dN
where a is in meter and DKI in MPa m . The centre crack in a large plate, initially of length
2 a = 8 mm, grows to 2a = 10 mm in 2000 load cycles when a constant amplitude fluctuating
load is applied with smax = 180 MPa and smin = 100 MPa. Determine the life of the
component beyond 2a = 10 mm if the same amplitude load continues on the component
and the maximum allowable crack length in the damage tolerant design is 2 a = 50 mm.
5. To find material constants for Paris law, it was found that an already nucleated centre
crack grows from 2 a = 5.6 mm to 2 a = 7 mm in 10,000 cycles of a constant amplitude load.
When the same load is continued, the crack grows from 2 a = 32 mm to 2 a = 36.8 mm in
1400 cycles. If smax = 180 MPa and smin = 90 MPa, find the constants C and m.
6. A centre-crack in a large plate is detected at 2 a = 5.0 mm when the plate is subjected to
fluctuating load of smax = 180 MPa and smin = 80 MPa. After 360 ¥ 103 cycles it grows to
2 a = 14.2 mm. Additional 180 ¥ 10 3 cycles are required to increase the crack length to
2 a = 37.4 mm. Determine elastic constants of the Paris law. (From the two equations,
eliminate C and determine m by trial and error.)
7. An edge crack in a large plate is subjected to a constant amplitude fatigue load with smax = 150 MPa
and smin = 10 MPa. Calculate the life of the component if initial crack length is 2.4 mm and
KIc = 80 MPa m . Also, find the life if correction for crack closure is made and U = 0.5 + 0.4
R. (The material constants of the Paris law are C = 6.0 ¥ 10–12 (MPa)–3 (m)–1/2, and m = 3.)
8. On a large plate, used as a critical component of a machine, amplitude of the fatigue load
shifts several times in a sequence of every 1000 cycles as shown in Fig. 9.22. The material
160
s, MPa
120
80
40
0
400
800
1200
1600
2000
N
Sequence 1
Sequence 2
Fig. 9.22 The figure of Problem 8
216
Elements of Fracture Mechanics
follows the Paris law with C = 2.2 ¥ 10–12 (MPa)3.4 (m)–0.7, m = 3.4. Determine how many
load sequences are needed to cause the failure if the initial crack of 2 a = 7.2 mm is detected
near the centre of the plate and KIc = 80 MPa.
9. A machine is used only for the first five days of a year and it is stored in an open ground
in the rest of the year where it is subjected to a corrosive atmosphere. A crucial component
of the machine is made of an alloy steel and its critical SIF is known as 60 MPa m . Just
before using the machine, a centre crack of length 2 a = 4 mm has been detected in the
component. During the operation of the machine the component is subjected to a fatigue
load of frequency 24 cycles per minute with its maximum stress of 120 MPa and minimum
stress zero. The material of the component follows the Paris law of fatigue growth given
by Eq. (9.3). Make an estimate of the remaining life of the component if the crack growth
rate under the corrosive conditions is 1.5 ¥ 10–7 mm/s. For the estimate to be conservative,
neglect the nucleation time for both fatigue and corrosive growth. Also, estimate the life
of the component if the machine is stored in a non-corrosive environment.
REFERENCES
9.1 Ellyin, Fernard (1997), Fatigue Damage, Crack Growth and Life Prediction. Chapman and
Hall, London.
9.2 Suresh, S. (1991). Fatigue of Materials, Cambridge University Press. Cambridge.
9.3 Barsom, J.M. and Rolfe, S.T. (1987). Fracture and Fatigue Control in Structures–Applications
of Fracture Mechanics, Prentice-Hall, Inc., Englewood Cliffs, New Jersey.
9.4 Pook, L.P. and Smith, R.A. (1979). Fracture Mechanics: Current Status, Future Prospect,
Proceeding of Conference, Cambridge. Editor: Smith, R.A. pp. 29–67.
9.5 Barsom, J.M. and McNicol, R.C. (1974). Effect of Stress Concentration on Fatigue Crack
Initiation in HY-130 Steel, ASTM STP 559. American Society for Testing and Materials,
Philadelphia, pp. 183–204.
9.6 Ashby, M.F. and Jones D.R.H (1980). Engineering Materials: An Introduction to Properties and
Applications, Pergamon Press, Oxford.
9.7 Forman, R.G., Kearney, V.E. and Engle, R.M. (1967). “Numerical Analysis of Crack
Propagation in Cyclic-Loaded Structure”, Journal of Basic Engineering, Transactions of ASME
89, pp. 459–464.
9.8 Erdogan, F. and Ratwani, M. (1970). “Fatigue and Fracture of Cylindrical Shells Containing
a Circumferential Crack”, International Journal of Fracture Mechanics, 6, pp. 379–392.
9.9 Gdoutos, E.E. (2005). Fracture Mechanics– An Introduction, Springer.
9.10 Broek, D. (1982). Elementary Engineering Fracture Mechanics, Martinus Nijhoff Publishers,
The Hague.
9.11 Meguid, S.A. (1989). Engineering Fracture Mechanics, Elsevier Applied Science, London.
9.12 Elber. W. (1976). “Equivalent Constant-Amplitude Concept for Crack Growth under
Spectrum Loading”, Fatigue Crack Growth under Spectrum Loads, ASTM STP 595, American
Society for Testing and Materials, Philadelphia, pp. 236–250.
9.13 Schijve, J. (1981). “Some Formulae for the Crack Opening Stress Level”, Engineering Fracture
Mechanics, 14, pp. 461–465.
Fatigue Failure and Environment-assisted Fracture
217
9.14 Broek, David (1988). The Practical Use of Fracture Mechanics, Kluwer Academic Publishers,
Dordrecht.
9.15 Hertzberg, Richard W. (1989). Deformation and Fracture Mechanics of Engineering Materials,
John Wiley & Sons, New York.
9.16 Kaesche, Helmut (2003). Corrosion of Metals: Physicochemical Principles and Current Problems,
Springer Verlag.
9.17 Janssen, M., Zuidema, J. and Wanhill,R. J. H. (2004), Spon Press, London.
9.18 Knott, J. F. (1973), Fundamentals of Fracture Mechanics, John Wiley & Sons, New York.
9.19 Raghavan, V. (1988), Material Science and Engineering, Prentice-Hall of India Pvt. Ltd.,
New Delhi.
9.20 Ramesh, K. (2007). e-Book on Engineering Fracture Mechanics, IIT Madras,
URL: http//apm.iitm.ac.in/smlab/kramesh/book_4.htm
Chapter
10
Finite Element Analysis
of Cracks in Solids*
Science studies what is, engineering creates what never has been.
von Karman
10.1 FINITE ELEMENT METHOD
Finite Element Method (FEM) is one of the numerical methods to obtain an approximate solution
to many of the fracture mechanics problems. This method has become very popular with the
availability of powerful computers. In the finite element method the domain of the problems is
discretized into a number of subdomains, called finite elements which are connected with one
another at points known as nodes. The variables of the problem, such as displacements,
temperature, velocity, etc., are approximated piecewise, so that they are represented in each
element by simple polynomials. The coefficients of the polynomial equivalently expressed as nodal
values of the variable are determined such that the governing equations and boundary conditions
are satisfied in the best possible manner. The approximation procedure may be a variational
method (such as minimization of potential energy) or a weighted residual approach [10.1, 10.2,
10.3]. In general, the solution becomes more and more accurate by choosing smaller elements and
more number of nodal variables. The necessary and sufficient conditions on the nodal variables
and the shape functions to assure the convergence are well laid out in FEM theory.
To briefly explain the methodology let us consider a two-dimensional stress analysis problem in
which the domain is divided into elements as shown in Fig. 10.1. Here the displacement variable
f( e) ( x, y ) in each element domain W(e) may be defined as a polynomial,
f ( e ) ( x , y ) = a1 + a2 x + a3 y + L + an yn
Equivalently, it can be expressed in terms of nodal values as
f (e) (x, y) = N1(x, y)f1 + N2(x, y)f 2 + L + Nn(x, y)fn
* Contributed by Prof. N.N. Kishore, Department of Mechanical Engineering, IIT Kanpur.
Finite Element Analysis of Cracks in Solids
219
Ti
Applied
traction
Ti
y
x
Displacement
boundary condition
Fig. 10.1 Two-dimensional body and finite element discretization
where f 1, f 2, …, f n, are the values of the function f at the n-nodes in the element, and N1(x, y),
N2(x, y) …, Nn(x, y) are known as interpolation functions, trial functions or shape functions. This
recasting of f (e)(x, y) in terms of nodal variables facilitates the automatic satisfaction of continuity
conditions at element boundaries.
For example, in a triangular element with nodes (i, j, k) at the three vertices, the displacement (u, v)
within the element can be written as (Fig. 10.2)
Pi
k
i
j
y
x
Fig. 10.2 Two-dimensional stress analysis using triangular elements
220
Elements of Fracture Mechanics
u ( x , y ) = N i (x , y ) u i + N j (x , y ) u j + N k (x , y ) u k
v ( x , y ) = N i (x , y ) v i + N j (x , y ) v j + N k (x , y ) v k
where ui , vi , u j , v j , uk , v k , are the nodal displacement values and N i , N j , N k are linear shape
functions. The strain components can therefore be expressed as
ex =
=
ey =
=
g xy =
=
∂u
∂x
∂N j
∂N i
∂N k
ui +
uj +
uk
∂x
∂x
∂x
∂v
∂y
∂N j
∂N i
∂N k
vi +
vj +
vk
∂y
∂y
∂y
∂u ∂v
+
∂y ∂x
∂N j
∂N j
∂Ni
∂N k
∂N
∂N k
ui +
uj +
u k + i vi +
vj +
vk
∂y
∂y
∂y
∂x
∂x
∂x
The above relations between strain components and the nodal displacements can be written in
the matrix form as:
È ∂N
Í i
x
Ï ex ¸ Í ∂
Ô Ô Í
Ì ey ˝ = Í 0
Ôg Ô Í
Ó xy ˛ Í
Í ∂N i
ÍÎ ∂y
0
∂N i
∂y
∂N i
∂x
∂N j
∂x
0
0
∂N j
∂y
∂N j
∂N j
∂y
∂x
∂N k
∂x
0
∂N k
∂y
˘ Ï ui ¸
0 ˙Ô Ô
˙ Ô vi Ô
∂N k ˙ Ô u j Ô
˙Ì ˝
∂y ˙ Ô v j Ô
˙
∂N k ˙ Ô u k Ô
Ô Ô
∂x ˙˚ Ó v k ˛
which in short form can be written as:
[e ]
= [ B ] {u( e ) }
The stress-strain relation, known as constitutive relation, for a linear isotropic material in plane
stress can be written as
Ïs x ¸
Ô Ô
E
Ìs y ˝ =
1-v 2
Ôt Ô
xy
Ó ˛
È
˘
Í1 v
0 ˙Ïe x ¸
Í
˙Ô
Ô
0 ˙Ì ey ˝
Ív 1
Í
1 - v ˙ ÔÓg xy Ô˛
Í0 0
˙
2 ˚
Î
Finite Element Analysis of Cracks in Solids
221
or in short,
[s ]
= [ D] {e }
By using minimization of potential energy of the body or virtual work principle, we can drive
the element stiffness matrix as
Ú [B] [D ] [B] h dA
T
where h is the thickness of the plate, and the integration is done over the area of the element
domain. These element stiffness matrices can be assembled to form global stiffness matrix [K]. The
applied traction in the form of distributed load can be transformed into nodal loads {P}. Thus, the
finite element equations can be finally written as global equations,
[K] {u} = {P}
In addition to the application of the load, it is also necessary to apply a minimum number of
displacement boundary conditions to remove the rigid body degrees of freedom and make the
finite element equations solvable. Having solved for global nodal displacements, the stress at any
point in any element can be evaluated using the stress and nodal displacement relations.
4-node quadrilateral
A
(e)
3-node triangle
ÈÎ K ( e ) ˘˚ =
8-node isoparametric
12-node isoparametric
x
Fig. 10.3 Typical two-dimensional elements
12-node isoparametric
4-node
quadrilateral
3-node triangle
y
8-node isoparametric
The above procedure for 3-node triangular element can be applied to other elements, such as
4-node rectangle, or 8-node isoparametric elements (Fig. 10.3). As these elements have higher
order interpolation functions, they will yield more accurate results. The isoparametric elements
can also model elements with curved edges.
222
Elements of Fracture Mechanics
For example, 12-node isoparametric elements have the capability to model quadratic variation
of stresses and strains. However, they require mapping and numerical integration technique to
evaluate element matrices (Fig. 10.4). In this element, the geometry is described by
y
7
8
9
10
6
11
5
t
10
9
12
8
1
7
11
6
12
5
2
3
s
1
2
3
x
4
Fig. 10.4 Mapping of 12-node element
12
x=
 Ni x i
i =1
12
y=
 Ni yi
i =1
and displacements are given by
12
u=
 N (s, t )
i
ui
 N (s, t )
vi
i =1
12
v=
i
i =1
where
1
(1 - t )
32
9
(1 - t)
N2 =
32
9
(1 - t)
N3 =
32
1
(1 - t )
N4 =
32
N1 =
4
(1 - s) ( -10 + 9s2 + 9t 2 )
(1 - s )
2
(1 - 3 s)
(1 - s ) (1 + 3 s)
2
(1 + s) (-10 + 9s 2 + 9t 2 )
Finite Element Analysis of Cracks in Solids
N5 =
N6 =
N7 =
N8 =
N9 =
N10 =
N11 =
N12 =
9
(1 + s)
32
9
(1 + s)
32
1
(1 + s)
32
9
(1 + t)
32
9
(1 + t)
32
1
(1 + t)
32
9
(1 - s)
32
9
(1 - s )
32
223
(1 - t 2 ) (1 - 3t )
(1 - t 2 ) (1 + 3t )
(1 + t ) ( -10 + 9s 2 + 9t 2 )
(1 - s ) (1 + 3 s)
2
(1 - s ) (1 - 3 s)
2
(1 - s) (-10 + 9s 2 + 9t 2 )
(1 - t ) (1 + 3t)
2
(1 - t 2 )
(1 - 3t )
10.2 DIRECT M ETHODS TO D ETERMINE FRACTURE PARAMETERS
The stress and strain fields in 2-D crack problems can be determined, in general, by using the
triangular, quadrilateral or isoparametric elements. The elements may be 3-node or 6-node
triangles, 4-node quadrilaterals or 8-node isoparametric elements as shown in Fig. 10.3. Out of all
these the 8-node isoparametric elements can model complex geometries and can have a quadratic
interpolation of displacement variation.
Making use of the expressions for the displacements and the stresses near the crack tip, we can
use the value of sij at any point near the crack tip to determine the value of KI, KII as follows:
ui = KI
s ij (r , q ) =
r I
r II
r III
fi (q ) + KII
fi (q ) + KIII
fi (q ) ,
2p
2p
2p
i = 1, º, 3
K II
K III
KI
g ijI (q ) +
g ijII (q ) +
g ijIII (q )
2p r
2p r
2p r
where fiI , fiII , fiIII and gijI , gijII , gijIII are functions of q.
For example, for a Mode I case, we know the stress, strain or displacement field for plane strain
as follows:
KI
qÊ
q
3q ˆ
cos Á 1 - sin sin ˜ fs
sx =
Ë
2
2
2¯
2p r
sy =
txy =
KI
2p r
KI
qÊ
q
3q ˆ
cos Á 1 + sin sin ˜
2Ë
2
2¯
q
q
3q
sin cos cos
2
2
2
2p r
224
Elements of Fracture Mechanics
u = 2 (1 + v)
KI
E
r
qÊ
qˆ
cos Á 1 - 2v + sin 2 ˜
Ë
2p
2
2¯
KI r
qÊ
qˆ
sin Á 2 - 2v + cos2 ˜
E 2p
2Ë
2¯
Therefore, KI can be determined from the known stress, strain or displacement field as
v = 2 (1 + v)
KI = s ij
2p r
gijI (q )
or from displacement field as
ui 2p
r fiI (q )
KI =
Ideally, KI and KII determined from the stress value at any point near the crack tip should give
the correct KI and KII values. However, there are certain limitations of simple elements to represent
large stress gradients. The stress gradient near the crack tip is very high and, theoretically, stress
becomes singular at the crack tip. To represent such large stresses and stress gradients reasonably
well, we need to employ a very fine mesh near the crack tip. Watwood [10.4] made some studies of
center cracked specimen using triangular and rectangular elements (Fig. 10.5). He evaluated KI
s•
2a
s•
(a)
5
6
7
1
2
3
4
Crack tip
(c)
a
(b)
Fig. 10.5 (a) Geometry and load on centre cracked panel, (b) finite element idealization of
upper right hand quadrant by Watwood [10.4], 478 Elements, 478 nodal points, and
(c) blow up near the crack tip
Finite Element Analysis of Cracks in Solids
225
using the stress values in various elements in the neighborhood of the crack tip; they are listed in
Table 10.1. It is clear from Table 10.1 that KI ranges from 5 to 18.5 while the theoretical value is 5.82.
TABLE 10.1 KI determined through sx, sy or txy in various elements of centre cracked
specimen (Fig. 10.5)
Element No.
KI
1
sx
sy
t xy
2.37
12.5
8.4
2
3.4
5.5
18.5
3
2.6
5.8
5.05
4
2.28
6.5
4.2
5
-- -
7.2
6.1
6
-- -
5.8
3.85
7
1.56
6.1
5.62
Very near the crack tip the finite element solution tends to be inaccurate due to its inability to
model singular nature of stresses accurately. An improved evaluation of KI can be done from the plot
of KI vs. r. Thus, KI can be extrapolated to find KI at r = 0 for a fixed q (Fig. 10.6). Using this approach,
Chan et al. [10.5] determined that the KI value with coarse mesh (area of element is 3.1 ¥ 10–4) was
within an error of 11% and with fine mesh (area of element is 1.2 ¥ 10–4) the value was within an error
of 5% compared to Westergaard solution.
2.0
Extrapolated
1.8
Theoretical
a)
1.4
KI/(s
1.6
1.2
FEM
1.0
0.8
0.6
0
0.2
0.4
0.6
r
a
Fig. 10.6 Results of finite element calculations for infinite plate with central
crack, Chan et al. [10.5]
226
Elements of Fracture Mechanics
The drawback of the above mentioned direct methods is that they require very fine mesh in the
vicinity of the crack tip which involves large amount of computational complexity.
10.3 INDIRECT METHODS TO DETERMINE FRACTURE PARAMETERS
There are several indirect methods which do not use crack tip stress formulae directly. They make
use of the knowledge of crack behavior in finite element analysis which will markedly improve the
accuracy of the results.
10.3.1 J-Integral Method
As given in Eq. (6.1), J-Integral is defined as
J=
Ê
Ú ÁËWdx
G
2
- Ti
∂ui ˆ
ds
∂x1 ˜¯
e
where W = Ú s ij de ij is the strain energy density and Ti is the traction. The contour G starts from one
0
crack surface and goes on to the other crack surface surrounding the crack tip (Fig. 6.2). This
integral value can be evaluated on a path which is slightly away from the crack tip. Thus, the stress
and the displacement values on the integration path G are not much affected by the modeling
inaccuracies of the crack tip stresses.
A typical path in a finite element analysis is shown in Fig. 10.7. The path is taken along the nodes
on the element’s edges. The strain energy density values at nodes are obtained as extrapolation of
the values at the Gauss points within the elements. The values of the traction vector Ti are the
external forces for the region enclosed by G as a free body. Chan et al. [10.5] applied this method to
the analysis of a compact test specimen. The resulting stress intensity factor KI is quite accurate and
has an error of only 3.5%.
G
Fig. 10.7 A rectangular elements mesh and path of integration G for the J-integral
Finite Element Analysis of Cracks in Solids
227
10.3.2 Energy Release Rate Method
This involves the evaluation of energy in the cracked body for two slightly different crack
configurations, say, a and a + Da. Let P and P + DP be the potential energies of the plate,
respectively (Fig. 10.8). Then, the energy release rate GI can be written as
G =-
DP
BD a
S
Extended
crack front
DI
S=0
Fig. 10.8 Growth of embedded crack
where B is the thickness of the plate. Watwood [10.4] applied this method to analyze a center
cracked panel of finite size. The GI can be written in terms of KI using the relation KI = GI E for a
plane stress case of linear isotropic material. The results are compared with those obtained by Isida
[10.6] in Table 10.2. It can be seen that the error is within 2%. This method can also be
applied to a 3-dimensional case to evaluate fracture parameters of the embedded cracks. An
average value of energy release rate G is given by (Fig 10.8):
Gav = -
DP
Ú Dl(s)ds
TABLE 10.2 KI obtained by FEM through energy release approach
a/W
K I /s
FEM
Difference
Theoretical*
0.188
4.86
4.96
2%
0.254
5.90
6.00
1.7%
0.317
6.90
7.01
1.6%
* Isida [10.6]
228
Elements of Fracture Mechanics
10.3.3 Stiffness Derivative Method
This technique evaluates the change in the potential energy DP in finite element analysis which
uses the change in stiffness of the plate DK for the two configurations of the crack. The difference in
the potential energies is given by:
(
)
1 T
u DKu + uT K - PT Du
2
where P are the nodal loads. This is suitable for 3-dimensional problems as the crack can advance
in multiple ways.
DP =
10.3.4 Singular Element Method
All the mentioned approaches discussed so far use conventional elements which either can model
constant stress or linear stress variation within each element. Modeling of large stress gradients
near the crack tip can be improved significantly by the employment of a very fine mesh. But they
are inadequate to model the theoretically infinite stress at the crack tip. Singular finite elements are
a special class of elements which have the exact interpolation functions to model the stress
singularity. In these elements, in addition to the usual nodal displacements as degree of freedom,
there will be displacement functions involving ( r1/2 ) terms. These are associated with additional
degrees of freedom directly giving the stress intensity factors KI and KII. Using Gifford and Hilton
[10.7] these additional terms are
r
2p
q
3q ˘
q
3q ˘ ¸
Ï
È
È
Ìcos q Í(2k - 1)cos - cos ˙ - sin q Í(2k + 1)sin - sin ˙ ˝
2
2
2
2 ˚˛
Î
˚
Î
Ó
u sI =
1
4m
usII =
1
r Ï
q
3q ˘
q
3q ˘ ¸
È
È
Ìcos q Í(2k + 3)sin + sin ˙ + sin q Í(2k - 3)cos + cos ˙ ˝
4 m 2p Ó
2
2˚
2
2 ˚˛
Î
Î
v sI =
1
4m
v sII =
1
r Ï
q
3q ˘
q
3q ˘ ¸
È
È
Ìsin q Í(2k + 3)sin + sin ˙ + cos q Í(2k - 3)cos + cos ˙ ˝
4 m 2p Ó
2
2˚
2
2 ˚˛
Î
Î
r
2p
q
3q ˘
q
3q ˘ ¸
Ï
È
È
Ìsin q Í(2k - 1)cos - cos ˙ + cos q Í(2k - 1)sin - sin ˙ ˝
2
2
2
2 ˚˛
Î
˚
Î
Ó
where m is shear modulus and
k = (3 – 4n )
= (3 – n )/(1 + n )
for plane strain
for plane stress.
The total displacement in the singular element is the summation of the regular expressions and
the terms derived due to singular terms are as follows:
12
u=
 Ni ui + KI usI +KII usII
i =1
Finite Element Analysis of Cracks in Solids
229
12
v=
 Ni vi + KI vsI +KII vsII
i =1
The singular elements are employed adjacent to the crack tip. Outside these singular elements,
the usual conventional elements are used.
Theoretical
KI = 2.00
10
10
1
3
s=1
1
Geometry and load
KI = 2.01
KI = 1.92
4 Elements
6 Elements
Idealizations of upper right hand quadrant
KI (Upper) = 2.05
KI (Upper) = 2.02
KI (Lower) = 2.05
KI (Lower) = 2.04
12 Elements
20 Elements
Idealizations of right hand side
Fig. 10.9 Geometry and idealizations for double edge notch tension
test specimen [10.7]
This singular element approach to a double edge notched specimen (Fig. 10.9) yields quite
accurate results using only a few elements [10.7]. With a 4-element mesh, the stress intensity factor
KI was evaluated as 1.92, and with a 6-element mesh the KI value was obtained as 2.01 while the
exact value is 2.00 [10.8]. For a single edge notch at an angle of 45° (Fig. 10.10) 4-element analysis
gave the values of KI and KII as 1.79 and 0.85 and 7-element mesh yielded 1.92 and 0.90 respectively
while the exact values are 1.86 and 0.88 [10.9].
230
Elements of Fracture Mechanics
KI = 1.79
KII = 0.85
s=1
45°
1.
0
2.5
2.5
2.5
4 Elements
KI = 1.92
KII = 0.90
7 Elements
KI = 1.88
KII = 0.94
10 Elements
KI = 1.89
KII = 0.88
18 Elements
Theoretical KI = 1.86, KII = 0.88
Fig. 10.10 Geometry, idealizations, and results for 45° slant crack in
tensile specimen [10.7]
10.3.5 Barsoum Element
An easier simulation of crack tip singularity can be achieved with the help of Quarter-Point element
technique which is known as Barsoum Element [10.10]. In this special method, the usual
isoparametric 6-node triangle or 8-node isoparametric quadrilateral elements are employed. As
shown in Fig. 10.11, mid-side nodes on the two adjacent sides are shifted towards the corner node to
the quarter point location. For these locations of the mid nodes, the Jacobian becomes singular at the
corner node, thus making displacement derivates infinite and consequently stresses and strains
become infinite as well. It can be shown that the variation of stresses along the two sides of the
elements is according to 1/ r .
On the other hand, if all the three nodes on the side of an 8-node quadrilateral element are
collapsed to one node (giving the same node number) then the stress (or strain) varies as 1/ r
along any radial line emanating from crack tip. To model the crack tip singularity the mesh
Finite Element Analysis of Cracks in Solids
7
6
Crack
5
4
8
1
Triangular element
5
6
7 8
231
2
3
Rectangular element
4
3
2
1
Quadrilateral element
collapsed to a triangle
Fig. 10.11 Quarter point elements
arrangement can be taken to be as shown in the Fig. 10.12. Observe that all the mid side nodes
adjacent to the crack tip are at quarter point locations. From the displacement field solution the
stress intensity factor KI , in a Mode I case, can be calculated as per the following relation,
KI =
2m
Ê 4v - v C ˆ
2p Á B
˜¯
Ë
k +1
L
where vB, vC are the displacement in the y direction behind the crack tip.
C
B
A
L
Fig. 10.12 Quarter point elements around a crack tip
It has been demonstrated that KI found by this method is within 2% of the theoretical solutions.
Accuracy of the finite element calculation can be improved if the neighboring elements are also
modeled to have the terms depicting the stresses for a crack with its tip outside the element.
Finally, finite element method can be applied to study fracture in the case of elastoplastic
materials and impact loads. Dynamic crack propagation can also be simulated well but the
propagation simulation procedures are still in progress.
232
Elements of Fracture Mechanics
REFERENCES
10.1 Cook, R.D., Malkus, D.S. and Plesha, M.E. (1989). Concepts and Applications of Finite Element
Analysis, John Wiley & Sons, New York.
10.2 Reddy, J. N. (1993). An Introduction to the Finite Element Method, McGraw-Hill, Inc.,
New York.
10.3 Zienkiewicz, O.C. and Taylor, R. L. (1989). The Finite Element Method, Vols. I and II,
McGraw-Hill, Inc., London.
10.4 Watwood Jr., V. B. (1969). “The Finite Element Method for Prediction of Crack Behaviour”,
Nuclear Engineering and Design, 11, pp. 323–332.
10.5 Chan, S. K., Tuba, I.S. and Wilson, W.K. (1970). “On the Finite Element Method in Linear
Fracture Mechanics,” Engineering Fracture Mechanics, 2, pp. 1–17.
10.6 Isida, M. (1955). “On the Tension of a Strip with a Central Elliptical Hole,” Transactions of
Japanese Society of Mechanical Engineering, 21, pp. 507–518.
10.7 Gifford, Jr. L. N. and Hilton, P. D. (1978). “Stress Intensity Factors by Enriched Finite
Elements,” Engineering Fracture Mechanics, 10, pp. 486-496.
10.8 Brown Jr. W.F. and Srawley, J.E. (1966). “Plane Strain Crack Toughness Testing of High
Strength Metallic Materials,” ASTM Special Technical Publication, 410.
10.9 Bowie, O.L. (1973). “Solutions of Plane Crack Problems by Mapping Techniques,”
Mechanics of Fractures, Vol. I, (Ed.) G. C. Sih, Noordhoff, Leyden, pp. 1-66.
10.10 Barsoum, R.S. (1976). “On the Use of Isoparametric Elements in Linear Fracture Mechanics,”
International Journal for Numeric Methods in Engineering, 10, pp. 25-38.
Chapter
Mixed Mode Crack
Initiation and Growth*
11
… I do not understand the reason why it is that the correct laws of physics seem to be
expressible in such a tremendous variety.
Richard Feynman
11.1 INTRODUCTION
In the previous chapters, we have studied crack growth under Mode I loading. Over the past many
years considerable research work has been devoted and fracture mechanics principle based designs
are developed using Mode I fracture theories. This may be mainly so, because Mode I failure
predominates for homogeneous isotropic materials. In the particular case of opening mode,
loading is applied in the direction normal to the crack faces and the crack grows parallel to the
faces, i.e., in a self-similar manner. But, in many practical situations, loading is of mixed mode
type, that is, combination of all the three Irwin’s modes—opening mode, in plane shear sliding
mode and out of plane tearing mode. For conservative fracture based design estimates, one needs
to characterize the crack under mixed mode loading.
Studies are carried in the mixed mode condition in order to find the crack extension direction,
critical load (or critical crack dimension) and stability of the crack path. Various models have been
proposed to characterize the mixed mode crack. Essentially, the models proposed are based either
on energy or stresses. Various models for initiation and growth of a crack subjected to Mode I and
Mode II loading are presented in this chapter within the scope of linear elastic fracture mechanics.
11.2 FRACTURE SURFACE
Fracture surface is a locus of points in KI–KII space where the combined action of KI and KII attains
a material dependent critical value. For a crack in a plane problem subjected to mixed mode
loading, the fracture surface can be taken in a general functional form as:
* Contributed by Prof. Raju Sethuraman, Department of Mechanical Engineering, IIT Madras.
234
Elements of Fracture Mechanics
f (KI, KII, KIc, KIIc) = 0
(11.1)
where KI and KII are Mode I and Mode II stress intensity factors, KIc and KIIc are critical values of KI
and KII. The function describing the fracture surface is commonly considered in a polynomial or
power law form and is expressed in a non-dimensional form as:
a
b
Ê KI ˆ Ê KII ˆ
ÁË K ˜¯ + ÁË K ˜¯ = 1
Ic
IIc
(11.2)
In the above equation the exponents a and b are material dependent constants. Equation (11.2) was
applied by Wu [11.1] for predicting the crack behavior in a mixed mode condition for composites.
He obtained the values of material constants a and b as equal to 1 and 2 respectively for the
behavior of a crack in wood. Different values have to be used for material constants a and b if one
needs to apply Eq. (11.2) for various composites.
Based on energy principles fracture surface function can be assumed to be quadratic in KI and KII.
It can be expressed as:
C 11K I2 + 2C 12 K I K II + C 22 K II2 = C
(11.3)
where C11, C12 and C22 are material dependent constants to be evaluated through experiments and
C is a real number constant.
Experimental results observed by Erdogan and Sih [11.2] follow ellipse like distribution as
shown in Fig. 11.1. This kind of distribution is a subset of the form given in the Eq. (11.3). Later,
Broek [11.3] proposed a model which takes the following form:
2
2
Ê KI ˆ Ê KII ˆ
ÁË K ˜¯ + ÁË K ˜¯ = 1
Ic
IIc
(11.4)
KII
KIIC
0
0
KI
KIc
Fig. 11.1 A typical failure surface based on experiment [11.2]
It was observed by Broek that the failure surface given by Eq. (11.4) closely approximates the
fracture behavior of mixed mode for most of the engineering materials.
11.3 MIXED M ODE C RACK PROPAGATION C RITERIA
In this section the following criteria are presented:
Mixed Mode Crack Initiation and Growth
235
∑ Modified Griffith Criterion
∑ Maximum Tangential Stress (MTS) Criterion
∑ Strain Energy Density (SED) Criterion.
11.3.1 Modified Griffith Criterion
In the Modified Griffith Criterion, the concept of energy balance is extended to include energy
release rates associated with all the modes. Total energy release rate for a crack in a plate subjected
to Mode I and Mode II loading is given as:
G = GI + GII
(11.5a)
where
GI = a
K 2I
E
(11.5b)
GII = a
K 2II
E
(11.5c)
a=1
= 1 – v2
for plane stress
for plane strain
According to this criterion, crack extension will occur in the direction where total energy release
rate is maximum and the extension will occur when the maximum energy release rate reaches a
critical value. The critical value depends on the material considered.
Crack Extension Direction
A crack in a plate with an inclination of b degree with ordinate subjected to remote loading s0 is
shown in Fig. 11.2. A polar coordinate system is considered at the crack tip. For various
hypothetical crack extension directions, q ( -p < q < p ) , the associated strain energy release rates
Gq is evaluated and is plotted as a function of q as shown in Fig. 11.3. The crack extension direction
qc corresponding to the G qmax is obtained by maximizing the Gq using
s0
X2
X ¢2
H
r
q
b
2a
X 1¢
q
c
X1
qc
s0
Fig. 11.2 Mixed mode crack subjected to remote loading s0
236
Elements of Fracture Mechanics
∂Gq
= 0
∂q
and
∂2 Gq
∂q 2
<0
(11.6)
max
Gq
Gq
0
qc
-p
0
q
p
Fig. 11.3 Variations of energy release rate with crack extension direction
Critical Condition
Crack extension will occur when G qmax reaches a critical value of strain energy release rate Gc , that is,
G qmax ≥ Gc
(11.7)
The material dependent critical value Gc is obtained from pure Mode I loading. Employing
Eq. [11.5(b)], we obtain
Gc =
a KIc2
E
where KIc is the critical Mode I stress intensity factor.
(11.8)
11.3.2 Maximum Tangential Stress Criterion
Maximum Tangential Stress Criterion (MTS) was proposed by Erdogan and Sih [11.2] based on a
certain component of stress state reaching a critical condition. Consider a crack subjected to Mode I
and Mode II loading as shown in Fig. 11.2. Stress components in the vicinity of crack tip (Point H)
using polar coordinate system are expressed as:
rr
= KIf
11
(r , q ) + K II f 12 (r , q )
(11.9)
qq
= KIf
21
(r , q ) + K II f 22 (r , q )
(11.10)
t rq = KI f
31
s
s
(r , q ) + KII f 32(r , q )
where the coordinate dependent functions f ij (r,q ) are given as
f11 (r ,q ) =
1
2p r
q 1
3q ˆ
Ê5
ÁË cos - cos ˜¯
4
2 4
2
(11.11)
Mixed Mode Crack Initiation and Growth
f12 (r ,q ) =
1 Ê 5
q 3
3q ˆ
ÁË - sin + sin ˜¯
4
2
4
2
2p r
f21 (r ,q ) =
1
2p r
f22 (r ,q ) =
1 Ê 3
q 3
3q ˆ
ÁË - sin - sin ˜¯
4
2 4
2
2p r
f31 (r ,q ) =
1
2p r
q 1
3q ˆ
Ê1
ÁË sin + sin ˜¯
4
2 4
2
f32 (r ,q ) =
1
2p r
q 3
3q ˆ
Ê1
ÁË cos + cos ˜¯
4
2 4
2
237
q 1
3q ˆ
Ê3
ÁË cos + cos ˜¯
4
2 4
2
According to MTS criterion, crack extension will occur in the direction where tangential stress
component s qq at an infinitesimal radial distance ro from the crack tip is maximum and the
extension will take place when the maximum tangential stress reaches a critical value which is a
material dependent parameter.
Crack Extension Direction
Consider a crack in a mixed mode loading condition as shown in Fig. 11.2. Around the crack tip an
infinitesimal circle with radius ro is constructed such that ro/a << 1 (Fig. 11.4). Along the
circumference of the circle, s qq is maximized and the corresponding qc is evaluated using
∂ s qq
=0
(11.12a)
∂ 2s qq
< 0.
∂q 2
(11.12b)
∂q
sqq
trq
srr
r0
a
qc
max
sqq
r0
a << 1
srr
Fig. 11.4 Crack extension direction for MTS criterion
238
Elements of Fracture Mechanics
Substituting f21 (r, q ) and f22 (r, q ) in Eq. (11.10), we obtain the expression for s qq as
s qq =
KI Ê
3KII Ê
3q ˆ
3q ˆ
q
q
ÁË 3cos + cos ˜¯ ÁË sin + sin ˜¯
2
2
2
2
4 2p r
4 2p r
(11.13)
Applying the condition of Eq. (11.12a) in the above equation and replacing q by crack extension
direction qc, we have
q
q
3q c ˆ
3q ˆ
Ê
Ê
+ KII Á cos c + 3cos c ˜ = 0
KI Á sin c + sin
Ë
Ë
2
2 ˜¯
2
2 ¯
(11.14)
This expression can further be simplified by trigonometric manipulation to:
KI sin q c + KII ( 3 cos q c - 1) = 0
(11.15)
Note here that the stress components srr and sqq are becoming principal stresses in the direction of
qc on the considered circle. This can be verified by substituting expressions of f13 (r, q ) and f23 (r, q ) in
Eq. (11.11) to have
qc
q
3q ˆ
3q ˆ ˘
È Ê
Ê
+ sin c ˜ + KII Á cos c + cos c ˜ ˙
ÍKI ÁË sin
¯
Ë
2
2
2
2 ¯˚
Î
Using Eq. (11.14) the above equation gives
t rq =
1
4 2p r
(11.16)
t rq = 0
Critical Condition
max
To obtain s qq
, the maximum value of sqq corresponding to crack extension direction, we replace
q by q c in Eq. (11.13). Also, we manipulate trigonometric relations to obtain:
max
=
s qq
q
KI
3
cos 3 c 2 2
2p r0
q
KII
cos c sin q c
2
2p r0
(11.17)
max
Crack extension occurs when s qq
reaches a critical value sc which is a material dependent
constant; sc is usually obtained from pure Mode I loading where qc = 0 and KI = KIc, that is,
sc =
K Ic
(11.18)
2p r0
For obtaining failure surface independent of the distance r0 , we substitute Eq. (11.18) in Eq. (11.17)
to have
KI cos 3
qc 3
q
- KII cos c sin q c = KIc
2 2
2
Fracture surface based on Eq. (11.19) is shown in Fig. 11.5.
(11.19)
Mixed Mode Crack Initiation and Growth
239
KII /KIc
3
2
0
0
KI /KIc
1.0
Fig. 11.5 Fracture surface based on maximum tangential stress criterion
Example 11.1 Consider an infinite plate with a center crack of length 2a. Find crack extension
direction and critical condition for the crack extension using MTS criterion for (i) pure Mode I
loading as shown in Fig. 11.6(a), and (ii) pure Mode II loading as shown in Fig. 11.6(b). The critical
stress intensity factor in Mode I, determined experimentally, is KIc.
s0
t0
2a
2a
s0
(a)
(b)
Fig. 11.6 Figures of Examples 11.1 and 11.2 of centre crack plane problems subjected to
(a) pure Mode I loading and (b) pure Mode II loading
Solution: (i) Invoking Eq. (11.15) for KII , we have
KI sin q c = 0
yielding
qc = 0
The critical condition for crack extension is obtained by substituting qc in Eq. (11.19) to have
KI = KIc
which, of course, is the expected value.
240
Elements of Fracture Mechanics
(ii) Invoking Eq. (11.15), for KI = 0, we have
KII ( 3cos q c - 1) = 0
yielding
q c = cos-1
1
3
There are two solutions for -p < q < p ,
q c = 70.5°
q c = – 70.5°
The second solution (q c = - 70.5∞) is considered because it satisfies Inequality (11.12b). The critical
condition for crack extension is obtained by substituting q c = -70.5 o in Eq. (11.19) to have
KII = -
KIc
2
3
KIc
=
3 cos (q c /2) sin q c
2
11.3.3 Strain Energy Density Criterion
Based on energy principles, Sih [11.4, 11.6] proposed Strain Energy Density [SED] criterion.
Consider a crack subjected to Modes I and II loading as shown in Fig. 11.2. Total strain energy
U can be obtained from stress and strain field as:
U=
Ú
V
Èe ij
˘
Í Ú s ij de ij ˙ dV
ÎÍ 0
˚˙
(11.20)
where V is a region inside the body. The SED function W is given as
dU
W=
=
dV
eij
Ú s ij de ij
(11.21)
0
For plane linear elasticity problems W can be written in the following form in terms of stress
components
(1 + v ) È k + 1
W=
(11.22)
(s 11 + s 22 )2 - 2 s 11s 22 - t 122 ˘˙
2E ÍÎ 4
˚
where
k = 3 - 4v
(for plane strain)
= (3 - v )/(1 + v )
(for plane stress)
(
)
The Cartesian stress components in the vicinity of crack tip in terms of polar coordinate system
are given as:
s11 = KI f11 ( r , q ) + KII f12 ( r , q )
(11.23)
s22 = KI f21 ( r, q ) + KII f22 ( r, q )
(11.24)
t12 = - KI f31 ( r , q ) + KII f32 ( r , q )
(11.25)
Mixed Mode Crack Initiation and Growth
241
where the coordinate dependent functions fij (r,q ) are given as
1
qÊ
q
3q ˆ
cos Á 1 - sin sin ˜
2Ë
2
2¯
2p r
f11 (r ,q ) =
f12 (r ,q ) = -
qÊ
q
1
3q ˆ
sin Á 2 + cos cos ˜
2Ë
2
2¯
2p r
f21 (r ,q ) =
qÊ
q
1
3q ˆ
cos Á 1 + sin sin ˜
2Ë
2
2¯
2p r
f22 (r ,q ) =
1
q
q
3q
sin cos cos
2
2
2
2p r
f31 (r ,q ) =
1
q
q
3q
sin cos cos
2
2
2
2p r
f32 (r ,q ) =
qÊ
q
1
3q ˆ
cos Á 1 - sin sin ˜ .
Ë
2
2
2 ¯
2p r
Substituting Eqs. (11.23)–(11.25) in Eq. (11.22), the SED function is obtained with some algebraic
manipulation in terms of stress intensity factor as:
W=
1
È g11 KI2 + 2 g12 KI KII + g 22 K II2 ˘
˚
p rÎ
g11 =
1
(1 + cos q )(k - cos q )
16 m
g12 =
1
sin q [2 cos q - (k - 1)]
16 m
g22 =
1
È(k + 1)(1 - cos q ) + (1 + cos q )(3 cos q - 1)˘˚
16 m Î
(11.26)
where
m=
E
2 (1 + v )
Strain energy density function poses singularity of order one at the crack tip. Sih proposed a
strain energy density factor S in a quadratic form which is independent of the coordinate r ; it is
defined as:
(
)
S (q ) = g11 K I2 + 2 g12 KI KII + g22 KII2 /p
(11.27)
According to SED criterion, crack extension will occur in the direction of minimum strain energy
density S (q ) and the extension will occur when the S (q ) reaches a critical value Sc which is a
material dependent parameter.
242
Elements of Fracture Mechanics
Crack Extension Direction
Consider a crack in a mixed mode loading condition as shown in Fig. 11.2. Along the circumference
of a circle with radius r0 strain energy density is minimized and the corresponding qc is evaluated
using conditions:
∂W
= 0
∂q
and
∂ 2W
∂q 2
>0
(11.28)
or,
∂2 S
>0
∂q 2
Substituting Eq. (11.27) in Eq. (11.29), we obtain
∂S
= 0
∂q
and
(11.29)
ÈÎ2cos q - (k - 1)˚˘ sin q KI2 + 2 ÎÈ2cos2q - (k - 1) cos q ˚˘ KI KII +
ÈÎ(k - 1 - 6 cos q ) sin q ˘˚ KII2 = 0
(11.30)
ÈÎ 2cos 2q - (k - 1) cosq ˘˚ KI2 + 2 ÈÎ(k - 1) sin q - 4 sin 2q ˘˚ KI KII +
(11.31)
ÈÎ(k - 1) cosq - 6cos 2q ˚˘ KII2 > 0
By knowing values of the stress intensity factors for a given problem, crack extension direction
can be obtained from the solution to Eq. (11.30), subject to the condition given by Inequality (11.31).
Critical Condition
Crack extension will occur when minimum value of strain energy density function (Smin ) reaches
a critical value of strain energy density factor Sc. Thus, the condition is expressed as:
(Smin ) ≥
Sc
(11.32)
The material dependent critical value Sc is usually obtained by using Eq. (11.27) for a pure Mode I
loading with qc = 0° and KI = KIc as
Sc =
(1 + v ) (k - 1)
4p E
KI2c
(11.33)
Thus, Inequality (11.32) can be expressed as
Smin ≥
(1 + v ) (k - 1)
4p E
KI2c
(11.34)
for a crack to become critical.
Example 11.2 Consider again the infinite plate of Fig. 11.6 with a center crack of length 2a. Find
crack extension direction and critical condition for crack extension using SED criterion for (i) pure
Mode I loading and (ii) pure Mode II loading. The critical stress intensity factor in Mode I,
determined experimentally, is KIc and Poisson’s ratio of the material is v = 0.3.
Mixed Mode Crack Initiation and Growth
243
Solution: (i) Involving Eq. (11.30) for KII for pure Mode I, we obtain
ÈÎ 2cosq c - (k - 1)˚˘ sin q c = 0
This equation yields two solutions:
qc = 0
Ê k - 1ˆ
q c = cos-1 Á
Ë 2 ˜¯
Substituting qc = 0 in the left hand side of Inequality (11.31), we obtain
LHS = ÎÈ 2 - (k - 1)˘˚ KI2
(11.35)
where
k = 3 - 4v = 1.8
(for plane strain)
k = (3 - v)/(1 + v ) = 2.08
(for plane stress)
Equation (11.35) simplifies to
LHS = 1.2KI2 > 0
LHS =
0.92KI2
(for plane strain)
>0
(for plane stress)
Hence, Inequality 11.31 is satisfied. It can be shown that the second solution does not satisfy
Inequality 11.31. Therefore, qc = 0 is considered. To find the critical condition, we substitute qc = 0
in Eq. (11.27) to have:
(k - 1) 2
KI
Smin = g11 KI2 =
8p m
Smin is substituted in Inequality 11.34 to obtain:
(k - 1)
8p m
K I2 ≥
(1 + v ) (k - 1)
4p E
KI2c
which simplifies to yield the expected result KI ≥ KIc.
(ii) Invoking Eq. (11.30), for pure Mode II loading, we have:
[(k - 1) - 6cosq c ] sin q c
=0
yielding two solutions:
qc = 0
Ê k - 1ˆ
q c = cos-1 Á
Ë 6 ˜¯
Ê k - 1ˆ
In the above two solutions, q c = cos-1 Á
is considered because it satisfies the condition of
Ë 6 ˜¯
Inequality (11.31). For a plane strain case with v = 0.3,
k = 1.8
Ê 0.8 ˆ
q c = cos-1 Á
= - 82.3∞
Ë 6 ˜¯
244
Elements of Fracture Mechanics
Substituting q c in Eq. (11.27), we have:
1+v
-k 2 + 14k - 1 KII2 .
96 E
To find critical condition for crack extension in pure Mode II, we substitute Smin in Eq. (11.34) to
have:
Smin =
(
)
Ê 24 (k - 1) ˆ
KII ≥ Á 2
˜
Ë -k + 14 k - 1 ¯
1/2
K Ic
It is worth noting here that MTS criterion is same for plane stress and plane strain because it is
stress based. On the other hand, SED criterion depends on conditions of plane stress or plane strain
because it accounts for stresses as well as displacements.
11.4 AN E XAMPLE
OF
M IXED M ODE
In the previous section, all three criteria have been discussed and relevant equations and inequalities
have been developed. In this section, we consider an example of mixed mode which is solved by MTS
criterion as well as by SED criterion.
Example 11.3 Consider an infinite plate with a crack of length 2a = 80 mm, inclined at angle b with
the applied tensile stress s 0 as shown in Fig. 11.7. KIc of the material is known to be 40 MPa m, its
elastic constants are E = 200 GPa and v = 0.3, and the plate is subjected to plane strain. (i) Determine
initial crack extension direction using MTS and SED fracture criteria for b = 60°, (ii) find the applied
stress s 0 corresponding to the crack initiation using MTS and SED fracture criteria for b = 60° and,
(iii) determine relations qc vs. b and critical s 0 vs. b for both fracture criteria for b varying between
10° and 90°.
Solution: State of stress with respect to axes X¢1 and X¢2 is (Fig. 11.7)
¢
s 11
= s 0 cos2 b
¢
= s 0 sin 2 b
s 22
¢
= s 0 sin b cos b
τ12
t¢12
s0
X2
s¢22
X¢ 2
b
s¢11
X¢ 1
X¢ 2
X¢ 1
X1
s¢11 = s0 cos 2 b
s¢22 = s0 sin2 b
t ¢12 = s0 sin b cos b
Fig. 11.7 Angled crack subjected to remote uniaxial tensile load
Mixed Mode Crack Initiation and Growth
245
Stress component s 22
¢ loads the crack in pure Mode I and component t 12
¢ in pure Mode II,
whereas the effect of s 11
¢ can be neglected. Thus, we have
KI = s 0 p a sin 2 b
KII = s 0 p a sin b cos b
(i) To find initial crack extension direction for b = 60°:
3
KI = s 0 p a sin 2 60∞ = s 0 p a
4
3
s0 p a
4
Substituting KI and KII in Eq. (11.15) for applying MTS criterion, we obtain
KII = s 0 p a sin 60∞ cos 60∞ =
3 sin q c + ( 3 cos q c - 1) = 0
This equation can be solved by trial and error. It is convenient to make a small program on
computer. Out of the two roots within the range -p < q < p , the one that satisfies Inequality (11.12b) is
q c = – 43.2°
Similarly, to find qc for SED criterion, we substitute KI and KII in Eq. (11.30) for plane strain
(k = 1.8) to obtain:
3 ( cosq c - 0.4) sin q c + 2 3 ( cos 2q c - 0.4 cosq c ) + (0.4 - 3cosq c ) sin q c = 0
Through trial and error approach this equation yields:
q c = – 40.5°
(ii) To find critical applied stress s 0crit corresponding to crack initiation:
To determine s 0crit through MTS criterion we substitute KI and KII in Eq. (11.19) to have:
crit
s0
p ¥ 0.040 È
˘
3 3
Ê - 43.2∞ ˆ
3 Ê - 43.2∞ ˆ
cos Á
sin ( - 43.2∞) ˙ = 40
Í3 cos Á
˜
˜
Ë
Ë 2 ¯
4
2 ¯
2
ÎÍ
˚˙
yielding s 0crit = 111 MPa.
For finding s 0crit through SED criterion, we substitute KI, KII and qc in Eqs (11.27) and (11.33) to
have:
Smin =
(
0.008286 crit
s0
E
)
2
132.4
E
which, on substitution in Inequality 11.34, gives:
Sc =
s 0crit = 126.4 MPa
246
Elements of Fracture Mechanics
b
0°
10°
30°
50°
70°
90°
–20°
qc
MTS
–40°
SED
–60°
–80°
Fig. 11.8 Critical angle qc vs. crack orientations angle b of Example 11.3 for
MTS and SED fracture criteria
(iii) Similar to parts (i) and (ii), qc and s 0crit are determined for various values of angle b using the
computer program. Figure 11.8 shows the relation between crack extension direction qc and crack
inclination angle b. For the pure Mode I case of b = 90°, the critical angle qc turns out to be 0°,
representing, as expected, crack growth in self-similar manner. As b decreases from 90° and the
Mode II component starts increasing, deviation of the crack angle (qc) from the crack plane increases.
Figure 11.9 shows the dependence of critical applied stress s 0crit on the crack inclination angle b.
Obviously b = 90° is worst possible crack orientation angle which shows the lowest value of s 0crit .
600
s 0Crit (MPa)
500
SED
400
MTS
300
200
100
0
10°
30°
50°
70°
90°
b
Fig. 11.9 Relation between critical load s crit
0 and crack angle b of Example 11.3
Mixed Mode Crack Initiation and Growth
247
11.5 CRACK GROWTH
For LEFM analysis, one of the three criteria, presented in this chapter, may be chosen for predicting
the initiation of the crack extension. In comparison to Modified Griffith Criterion, MTS Criterion
and SED Criterion are more popular among investigators. Furthermore, MTS criterion is strictly
stress-based and its analysis does not depend on the condition of plane stress and plain strain. SED
criterion deals with strain energy and therefore conditions for predicting crack growth direction
and critical applied stress depend on whether the plate is subjected to plane strain or plane stress.
The path of a finite crack growth in a mixed mode case can be predicted by successive
application of the fracture criterion. When the given external loading situation reaches the critical
stage through a chosen criterion, the existing crack is extended in the predicted critical direction
through a small distance. For the newly extended crack configuration we apply the chosen criterion
again and criticality of the current (updated) loading configuration is checked. If the loading is
critical, the crack is extended and the above procedure is repeated. If the condition is subcritical,
the crack growth stops.
REFERENCES
11.1 Wu, E.M. (1967). “Application of Fracture Mechanics to Anisotropic Plates,” Journal of
Applied Mechanics, 34, pp. 967–974.
11.2 Erdogan, F. and Sih, G. C. (1963). “On Crack Extension in Plates under Plane Loading and
Transverse Shear, Transaction of ASME,” Journal of Basic Engineering, 85, pp. 519–527.
11.3 Broek, D. (1974). Elementary Engineering Fracture Mechanics, Noordhoff, Leiden.
11.4 Sih, G.C. (1973). “Methods of Analysis and Solutions of Crack Problems,” Mechanics of
Fracture, Vol. I, (Ed.) G. C. Sih, Noordhoff, Leiden.
11.5 Sih, G.C. (1973). “Some Basic Problems in Fracture Mechanics and New Concepts,”
Engineering Fracture Mechanics, 5, pp. 365–377.
11.6 Sih, G.C. (1974). “Strain Energy Density Factor Applied to Mixed Mode Crack Problems,”
International Journal of Fracture, 10, No. 3, pp 305–321.
Chapter
Crack Detection through
Non-Destructive Testing
12
Archimedes was given the task of determining if King Hiero's goldsmith was embezzling
gold during the manufacture of the king's crown…Baffled, Archimedes went to take a bath
and observed from the rise of the water that he could calculate the volume of water.
Allegedly, Archimedes went running through the street naked shouting,
"Eureka! Eureka!"
Wikipedia
12.1 INTRODUCTION
There are defects in the materials of real life components, which act as cracks. In fact,
manufacturing of a component is not ideal. There are always defects of several kinds like voids and
inclusions. Welding, a common technique to join components, is a very aggressive technique. It
generates various kinds of defects in the weldment and in the heat affected zones. Thus, all critical
components must be checked through non-destructive techniques (NDT) to detect potentially
dangerous cracks.
It is common practice to apply a factor-of-safety during the designing of a work-piece. The major
uncertainty is due to the presence of manufacturing defects. If the work-piece is scanned with an
NDT to rule out dangerous cracks, its integrity is assured and reliability is improved. The designer
can afford to choose a lower factor-of-safety. These days, NDT has been improved to the extent that
almost defect free components can be manufactured for crucial applications. However, a designer
should justify the cost involved.
Many process plant operators and machine users now perform scheduled inspection at a regular
interval rather than act after a failure. NDT is very effective in identifying those defects which may
grow and cause catastrophic failure in a component. These defects can either be repaired or else the
component itself can be replaced. This kind of quality assurance is found to be economical in the
long run.
Crack Detection through Non-Destructive Testing
249
In certain crucial components such as space vehicles, airplanes, nuclear plants, dams, etc., online
monitoring of defects is recommended. To achieve it, sophisticated technologies have been
developed; many of them use large data handling capabilities of modern computers. These days
many non-destructive tests are available, from very simple to very sophisticated. This chapter
discusses some of the commonly used non-destructive test methods. They are:
∑
∑
∑
∑
∑
Examination through human senses,
Liquid penetration method,
Ultrasonic testing,
Radiographic imaging,
Magnetic particle inspection.
12.2 EXAMINATION
THROUGH
HUMAN SENSES
In many cases, a defect in a component can be detected in an initial investigation using human
senses such as sight, smell, hearing, etc. Field technicians working with a machine usually know
the possible locations where a crack is most likely to grow and may become unstable. Also, these
people with some experience develop a keen and acute sense of observation.
12.2.1 Visual Inspection
A human eye is a marvelous device provided by nature. It is capable of detecting even minute
changes in the intensity of light. For most crack detections, an initial cleaning of the surface is
required. My friend, Vijay, is an expert on rolling mills. Once Vijay was consulted by a production
firm to determine why the rolled sheets of a rolling machine were no longer maintaining the
uniform thickness. I accompanied him to the site; Vijay looked at the rolling mill and then asked for
a cleaning cloth. The frame which houses the rollers is a crucial component. It was found to be oily
and dusty. Vijay cleaned the portion of the frame between the main rollers and observed a fatigue
crack growth. The crack had still not achieved the critical length required for catastrophic failure
but had decreased the stiffness of the frame considerably. The firm owner was advised to reinforce
the frame so that the crack could be arrested and the frame could regain its original stiffness.
Several optical aids are used to facilitate the detection of a crack. Most of these devices have
their own light source to properly illuminate the area under investigation. The simplest optical
device is a convex lens known as the magnifying lens. To investigate the surface for a crack the lens
is moved in or out to focus on the surface. A magnification of 2 to 3 usually facilitates crack
detection. To detect minor cracks or scratches, an optical microscope is used, usually of 10
magnifications. A microscope with a higher magnification of 100x, 500x or 1000x can also be used
but the depth of the field and the area of view are reduced considerably.
An endoscope is a more sophisticated optical device which has its own source of light so that the
surface can be illuminated and scanned for cracks. The surface is adequately illuminated by the light
carried through flexible optical fibers. Also, a subminiature closed circuit TV camera is installed at
the front portion of the endoscope. The pictures are then carried back with the help of another set of
optical fibers to a computer screen. Because of high optical efficiency of optical fibers, an endoscope
can view surfaces up to 4–5 meter away. Endoscopes are useful for a wide range of applications like
250
Elements of Fracture Mechanics
detecting cracks in the internal surfaces of boilers, pipes, reactors, and heat exchangers. For example,
once I visited a company making cylinders to store nitrogen and oxygen gases at very high pressures.
I found that the internal surface of each cylinder was being thoroughly checked with an endoscope.
12.2.2 Investigation through Hearing
Polymer composite laminates are initially investigated by a coin test. A metal coin (e.g. one rupee
coin) is tapped on the surface of a polymer composite component. The noise of the tap changes
wherever interlaminar separation is present. With no interlaminar damage, the sound waves go all
the way up to the rear end of the laminate and then return back, thus taking a long time to return.
This corresponds to a low frequency noise. When interlaminar cracks are present, some sound
waves are reflected from the surfaces of the interlaminar separation and return back early giving a
higher frequency noise. The difference in the frequencies can be quite easily detected. However, a
component passing the coin test is usually checked further through a rigorous NDT like ultrasonic
investigation before it is finally approved.
Up till very recent time, all axles of locomotive passenger compartments were tested on all big
stations during a long railway journey. Now, with the availability of more reliable components,
frequent checking is not required. The axles are inspected in the yard of destination station. A
trained person taps one end of the axel with a small hammer to hear the frequency. A higher
frequency note would cause alarm and the compartment with a defective axle is taken out of the
train-rack for a rigorous NDT. In short, appearance of a crack in a component does produce new
sound frequencies and a trained person can sense the danger.
12.2.3 Detection through Smell
One good example is the leakage in an LPG cooking stove. In reality, there is no smell in LPG but an
artificial smell is introduced so that if there is a leakage any where, a housewife can smell the gas
and be alarmed. In most cases, the damage occurs in the plastic pipe taking the gas from the
cylinder to the stove as it gets aged and cracked.
In making a high pressure cylinder, the design is usually made to have a leak before break. A
semi-elliptical crack extends mostly in depth from the internal surface towards external surface. A
stage comes when the semi-elliptical crack grows and becomes a through-the-thickness-crack. The
design is made such that the through-the-thickness-crack is not critical. However, there will be a
leakage of the gas or liquid and the crack is detected.
12.2.4 Other Simple Methods
I visited a steel plant which was making tubes as one of the several products. Each tube was tested
by first closing its ends and then filling it with a liquid (e.g., water). The pressure of the liquid was
increased to 30–40 % higher than the designed pressure. The pressure of the filled system was
monitored with a pressure gauge for a specified time (e.g., 15 minutes) to explore whether the
pressure decreases. A flow through a leak will make the pressure drop fast as the liquid does not
compress much and even a small loss of liquid will decrease the pressure quickly. Such pressure
difference tests are adopted for checking various systems. If a system stores high pressure gas, its
leakage is often detected by applying a soap water solution on the suspected place and watch the
bubbles come out.
Crack Detection through Non-Destructive Testing
251
12.3 LIQUID PENETRATION INSPECTION
Liquid penetration inspection is a simple technique for non-destructive testing and is one of the
most widely used techniques. It detects cracks which are exposed to a surface. The technique can
be used on the surface of a wide variety of materials. It is quite reliable and is capable of identifying
cracks of very small width, even cracks of just a few microns (µm).
12.3.1 Principle
An appropriate penetrant liquid with an excellent wetting capability is made to enter into the
cavity of a crack through the capillary action. It is worth noting here that widths of most cracks in
work-pieces of real life are so thin that cracks can not be observed through naked eyes. Figure
12.1(a) shows one such crack: dashed lines depict the existence of a crack which is invisible to
human eyes. When a penetrant is applied on the surface it starts entering the crack [Fig 12.1(b)].
The penetration into a very narrow crack cavity is slow and thus takes time, known as dwell time.
The penetrant contains a bright colour dye, usually red, to provide enough contrast for decorating
the crack.
Fig. 12.1 (a) A thin crack having an opening on the surface but invisible, (b) a penetrant is
spread on the surface which enters in to the cavity of the crack, (c) the penetrant is wiped
out but crack is still invisible and (d) spreading of developer brings colored penetrant
out of the crack cavity to make of crack look wider and visible
The excess penetrant is wiped off from the surface, leaving only the liquid within the crack
cavity. Although the crack is filled with a colored liquid [Fig. 12.1(c)], the crack can not be seen yet
in most cases because the crack mouth is very narrow (a few µm only).
Another liquid, a developer, is applied on the surface which is a good solvent for the penetrant
liquid. As a result, a part of the penetrant comes out of the crack cavity and diffuses into the
developer's layer on the surface [Fig.12.1 (d)]. In fact, the developer's layer works like a blotting
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Elements of Fracture Mechanics
paper. Thus, a colored line with a reasonable width appears at the mouth of the crack opening. The
colored line can now be easily observed by the investigator.
12.3.2 Procedure
Before applying the penetrant, the surface of the work-piece is made free of scales, flakes, paint,
dirt, grease, other chemicals, etc. If the surface is not cleaned, the penetrant will have difficulty
entering the crack cavity. However, abrasive cleaning methods should be avoided. Sand blasting
must not be done as the particles are likely to fill the crack opening. Usually, a good solvent is used
to clean the surface. Then, the surface should be dried so as to let the cleaning solvent evaporate
out and to facilitate the entering of the penetrant inside the crack.
A penetrant is usually a water based liquid containing a bright dye, usually red. It can be
petroleum based also. A supplier usually provides a kit of mutually compatible chemicals—a
penetrant, a developer and useful solvents. It is important that the penetrant and the other
chemicals do not react with the surface of the work-piece. Filling of the crack cavity through
capillary action takes time and, therefore, the layer of penetrant should be allowed to remain on
the surface for some time. The dwell time is usually 20 minutes or so.
At the end of the dwell time, the penetrant is cleaned from the surface. The surface should be
cleaned thoroughly. If the traces of the penetrant remain on the surface, they will be decorated too
along with the crack cavity when the developer is applied and the identification of a crack may
become difficult. The cleaning should not be overdone also, as it can cause the loss of the penetrant
from the cavity of the crack. For water-based penetrants, a simple wiping of the surface with a
cloth damped with water is suitable. Or the work-piece may be washed in water. For petroleum
based penetrants, oil or chlorine based solvents may be used.
A developer which is a good solvent of penetrant is applied on the surface. The developer is
usually a liquid but in certain cases a solid powder is also used. The penetrant takes time to come out
of the crack cavity and get soaked into the developer. This dwell time is approximately same as the
dwell time taken by the penetrant to enter into the crack. The sideway spread of the bright colored
dye makes the crack line look thicker. Then, the crack can be easily detected with naked eyes.
Chemicals used in dye penetration tests are usually sold in convenient aerosol cans. Thus, the
chemicals can be easily sprayed on the surface to be examined for crack detection. Because of the
low cost of chemicals and their availability in convenient aerosol spray cans, the dye penetration
technique is very popular.
12.3.3 Crack Observation
The crack should be observed in bright day light or in a well illuminated place. The wavelength of
red light lies between 0.63 and 0.76 µm which is adequate for detecting most cracks. It is important
to know that to identify a geometric feature, it has to be greater than or equal to l/2 where l is the
wavelength of the light source. An analogy is that when a horse crosses a field, he does not feel the
small pits or bumps on the surface of the ground. But, if a mouse runs over the ground he has to go
up and down with the unevenness of the surface. However, if the width of a crack is expected to be
quite small, a fluorescent dye is used in the penetrant in place of the red colour dye. The cracks
are then observed under dark conditions using ultraviolet (UV) rays whose wavelength is
substantially smaller (in the range of 0.35 to 0.4 µm).
Crack Detection through Non-Destructive Testing
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The dye penetrant examination can be performed on a variety of materials, ferrous and nonferrous metals, polymers, ceramics, and glass, etc. However, the technique is not suitable for porous
materials because pores that are opened to the surface will light up and it would be difficult to
identify the cracks.
For thin components, through-the-thickness-cracks are identified easily as the penetrant passes
from the front free surface to the rear free surface (Fig. 12.2).
Fig. 12.2 Detecting a through-the-thickness crack
To summarize, the liquid penetration inspection is inexpensive, simple to use, reliable and
versatile to inspect work-components of many kinds of materials. But it can detect only those
cracks which are exposed on the surface. For fully embedded cracks, other non-destructive
techniques should be used.
12.4 ULTRASONIC TESTING
Ultrasound tests in medical diagnostic centers are now quite popular and many of us have
undergone scanning of our body organs. Ultrasonic testing to detect crack in a hardware
component is based on the same principle. In fact, the ultrasonic tests were first developed for
hardware materials and later on modified to perform ultrasound diagnosis of human body.
Ultrasonic testing is non-invasive, simple, inexpensive and very versatile. The technique is capable
of detecting fully embedded cracks as well as surface cracks. Ultrasonic detectors are now made
very compact and can be easily taken to inspect defects in fields. Also, defects can be easily identified
in many kinds of materials like metals, polymers, wood, ceramics, polymer composites, etc.
12.4.1 Principle
The ultrasonic testing is similar to a search light being used to locate a thief. If there is no thief, the
light gets reflected from a wall across the street. In case, there is a thief, the light will be reflected
from the thief as well as from the wall.
In case of a search light, optical light waves emanate and detect objects, where as in ultrasonic
tests, stress waves (also known as the sound waves) propagate within the work-piece to be
investigated. When a stress wave propagates through a solid or a liquid, a particle vibrates from its
usual position and makes the neighboring particle move. Thus, the energy propagates from one
particle to another with sound wave velocity.
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There are several kinds of stress waves in a material but two prominent ones are, (i) longitudinal
or primary waves in which the particles move in the direction of energy propagation, and (ii) shear or
secondary waves in which particles move normal to the direction of energy propagation. Usually, the
velocity of longitudinal waves is designated by c1 and the velocity of shear waves by c2. They can be
determined with the help of following expressions
c1 =
E/ r
c2 =
G/ r
where E is theYoung's modulus, G the shear modulus and r the density. It is to be noted that c1 is
greater than c2 as E is larger than G. Majority of ultrasonic tests are done through primary waves but
secondary waves are also used for special applications. Primary wave velocity in steel is quite high
(about 5140 m/s).
In ultrasonic testing, a bunch of stress wave pulses (usually 3 to 9) of a high frequency is
introduced on the surface of a work-piece by a probe as shown in Fig. 12.3. The frequency of the
pulse-bunch is high, usually in the range of 0.5–15 MHz. This frequency is much higher than our
capability of hearing whose upper limit is only 0.02 MHz. Thus, ultrasonic testing is inaudible to
human ears. Most of the stress waves of a pulse-bunch reflect back from the rear face of a workpiece since there is air beyond it and very little energy is transmitted to air particles. In fact, for
waves propagating in solids and liquids, the air can be regarded as vacuum. The reflected pulses
are received by the probe and are quite weak in magnitude. The probe sends them to an instrument
for amplification and processing. If a defect exists in the work-piece, the pulse-bunch is reflected
from the defect with a shorter flight time.
Probe
(a)
Defect
Work-piece
(b)
Fig. 12.3 (a) A bunch of high frequency pulses, (b) a probe transmitting high frequency
pulses to a work-piece and receiving reflections from the rear surface and a defect
It is inconvenient to deal with a bunch of high frequency pulses. Therefore, pulses are rectified
and smoothened as shown in Fig. 12.4; the smoothened pulse is known as echo. The incident pulses
on the work-piece are also rectified and smoothened to have a primary echo (bang). Figure 12.5
shows the primary echo, the echo from the rear surface and the echo from a defect. Appearance of
an echo between the primary echo and the echo of the rear surface gives two important
informations, (a) there is a defect in the material, and (b) how deep is the defect. Figure 12.5 shows
Crack Detection through Non-Destructive Testing
255
an echo from a defect with round trip flight time of t1. Then, the distance of the defect from the
front surface is c1t1/2.
Rectified pulses
Echo
A bunch of high
frequency pulses
Fig. 12.4 A burst of high frequency pulses rectified and smoothened to have an echo
Primary echo
Echo from defect
Echo from
rear surface
t1
Fig. 12.5 Echoes recorded on an oscilloscope display or a computer
Ultrasonic testing requires high frequency of pulses because the defect size should be larger
than the half of the wavelength (l/2) of elastic waves. For example, wavelength of sound wave
propagation for 5 MHz probe is about 1 mm in steel. This kind of pulse frequency can detect flaws
having geometrical features large than 0.5 mm. If a probe of 1 MHz frequency is used on a steel
component, defects smaller than 2.5 mm will not be detected. On the other hand, wavelength of
1 MHz probe in a polymer like epoxy is close to 1.6 mm and defects of sizes smaller than 0.8 mm
are not detectable.
12.4.2 Equipment
There are mainly three major units of an ultrasonic test setup, (i) probe, (ii) pulse-echo instrument,
and (iii) display oscilloscope (or computer) as shown in Fig. 12.6. The probe is the most crucial unit
and will be taken up first.
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Work-piece
Probe
Scope display
or computer
Pulse-echo instrument
Fig. 12.6 Hardware elements of an Ultrasonic Detector
Probe
The transducer in the probe is a thin disk of a piezoelectric material which converts a voltage signal
into stress pulses. These stress pulses are transmitted to the surface of the work-piece. There are
several piezoelectric materials available but quartz is found to be appropriate for most
applications. The thin disk is cut from a single crystal of quartz at an appropriate angle (X-cut for
generating longitudinal waves). Figure 12.7 shows the schematic diagram of a probe. The thickness
of the quartz disk is precisely controlled so as to have its resonant frequency same as the desired
frequency of the probe. To apply voltage on the quartz transducer both faces of the thin quartz disk
are electroplated and electric wires are attached. At the front face of the quartz disk a wear resistant
plate is fixed as shown in the figure. A spike of high voltage and of a very short duration (ª10 ns) is
applied on the quartz transducer by the pulse-echo instrument. This makes the piezoelectric disk
vibrate at its resonant frequency. The vibrations in quartz transducer do not die fast and, therefore,
a damping material is applied on the rear face of the piezoelectric disk. The damping material
helps in suppressing the vibrations of the disk and thus, only a bunch of a few pulses passes into
the work-piece.
Damper
Quartz
transducer
Electrode
wires
Wear plate
Electrode
Fig. 12.7 Schematic diagram of a probe with quartz transducer, sound absorbing
damper and wear plate
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257
A kit of ultrasonic flaw detector comes with probes of different frequencies. Depending upon
the material to be tested, an appropriate probe is chosen. The wavelength of sound waves is usually
maintained between 1 and 10 mm within the work-piece.
The piezoelectric disk also works as a transducer to record the echoes coming back to it. The
stress of an echo develops voltage across the faces of the piezoelectric disk. The electric signal thus
generated is weak and needs to be amplified.
A material known as couplant is required to propagate the sound waves from the probe to the
surface of the work-piece. Usually, a viscous oil, grease, glycerin or water is used as couplant. The
couplant should be quite inert to avoid damage to the surfaces of the work-piece and the probe.
Pulse-Echo Instrument
As already stated, the pulse-echo instrument generates a spike of high voltage and of a very short
duration at a regular interval. This spike resonances the piezoelectric disk which, in turn, passes a
bunch of pulses into the specimen as the primary echo (main bang).
The sound wave inside the material is dissipated due to several mechanisms (to be discussed
subsequently). Once all the echoes of a main bang die, a new spike of identical nature is generated.
The new spike again triggers the oscilloscope and the echo response is repeated. This kind of
repetition at a regular interval creates a stable display on an oscilloscope or computer screen.
An amplifier in the pulse-echo instrument amplifies the weak signal of the echo received by the
probe. Also, the instrument filters the undesirable noise picked up by the probe before the echo is
displayed on an oscilloscope screen. Some professionals like to use a computer in place of
oscilloscope display. It has the advantages that a large number of test information can be stored, an
echo can be magnified and minute features can be studied.
Transmission Losses
As the pulse propagates in the material, there are transmission losses and the intensity of the stress
waves is attenuated because of several causes: (i) there is a divergence of stress waves which in
turn makes the echo weak,(ii) there is a scattering of stress waves due to inhomogeneities present
in real life material such as inclusions of second phase particles, voids, shrinkage cracks, grain
boundaries, etc., and (iii) there exist hysteresis which convert mechanical energy of sound waves
into heat. If the transmission losses are high, the echo from the rear surface of the work-piece will
be of poor quality. Losses of high frequency stress waves are more and therefore a high frequency
probe may not be appropriate for testing thick components. A low frequency probe can be used on
a thick work-piece as the transmission losses are low. However, it can not detect minute defects as
the wavelength of the sound wave is long. On the other hand, a high frequency probe can detect
defects of smaller sizes. An experienced or trained operator chooses a probe of a right frequency
for the given job.
Transmission losses are much higher in polymers in comparison to losses in metals. Then, a
single probe with pulse-echo capability may not work. In such situations, two probes are used, a
transducer transmitter and a transducer receiver, as shown in Fig 12.8. Most polymer composite
laminates are tested with the help of this arrangement.
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Transducer
transmitter
Work-piece
Transducer
receiver
Fig. 12.8 Two probes Ultrasonic Detector
12.4.3 Immersion Inspection
A component is completely immersed in water or other suitable liquids and tested for flaws. The
testing can be done through a transmitter-receiver transducers or through a single probe using
pulse-echo principle. The immersion method is convenient for testing a large number of workpieces. The method has an additional advantage that the couplant is uniform over the entire surface
of a work-piece.
From the geometric point of view, there are three kinds of inspections–A-Scan, B-Scan and
C-Scan. A-Scan investigates flaws at a point of the work-piece. If the probe is moved in a line and
echo response is continuously monitored it is called a B-Scan. C-Scan is useful in scanning the full
area of a component as the probe is moved first along a line, say parallel to x-axis, then, it is shifted
laterally by a small amount along y-axis and again scanned in x-direction. The procedure is carried
out to cover the entire area. Figure 12.9 shows a C-scan of an impacted polymer composite panel.
The empty space corresponds to the signals not received by the receiving transducer. In the
impacted panel, this corresponds to delamination caused by the impact induced damage.
y
x
Fig. 12.9 C-scan showing delamination in a polymer composite laminate
Once, I visited a company which was making large panels of polymer composites to be used in
airplanes. They had a setup specially made for detecting crack in a panel. In the set up, a jet of
water hit the left face of a panel which also transmitted ultrasonic waves to the panel. On the right
face of the panel, another jet of water was made to hit the panel in the same line. This water jet was
connected to a receiving transducer. The transmitting and receiving transducers moved in unison
to do the C-scan of the work-piece.
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259
The ultrasonic testing is very versatile and has been adopted for various applications by
devising special elements. For example, a work-piece can be scanned at an angle using an angular
probe whose transducer usually works on shear wave propagation. In short, ultrasonic test is a
widely used non-destructive technique. It is simple to use, relatively inexpensive, quick and
portable. The technique can detect surface as well as fully embedded defects.
12.5 RADIOGRAPHIC IMAGING
Working of radiographic imaging is similar to that of a camera with a flash light which we use in
daily life. The flash sends a predetermined amount of light and the reflected light from the objects
is recorded on a photographic film installed inside the camera. The film records the contrast of the
light reflected from the various objects in the view. In radiography, the difference is that the
recording film is separate from the radiographic source and is placed behind the work-piece.
In radiographic imaging, electromagnetic waves (X-rays or g -rays) of very short wavelength are
transmitted through a work-piece. If the material of the work-piece is not uniform, the transmission
of electromagnetic waves will be absorbed differently and there will be contrast on the recording
film. For example, if there is an embedded void in the work-piece, the electromagnetic waves will
not be absorbed while propagating through the void and thus reduction in the intensity will be
less. Consequently, the film will be exposed more on the portion that corresponds to void location.
X-rays or g -rays are both electromagnetic waves but are produced using different sources. Each
has its own strength and limitations (to be discussed subsequently). For radiographic imaging,
both waves behave in same manner to detect defects.
12.5.1 Contrast through Absorption Rate
X-rays or g -rays propagate with the velocity of light. Their frequencies are extremely high and
consequently their wavelengths are extremely small, usually ranging between 1 nm and 1 ¥ 10 –6
nm. They are capable of passing through any material because of the presence of spaces between
atoms. Also, an atom itself is spacious as its nucleus is quite small. Radiographic waves are like a
mouse which can go through small holes in a wall, whereas light waves with longer wavelength
are like a cat which cannot pass through.
X-rays or g - rays are progressively absorbed as they propagate in a material. The absorption rate
(attenuation) depends on the material being tested; it is high in a dense material. The progressive
absorption is the key to radiographic imaging. An inclusion or a void will affect the intensity of
radiation on the recording film, thus creating contrasts.
12.5.2 Imaging through X-rays
X-rays are produced in high vacuum within a tube made of glass or ceramic. When the cathode is
heated by passing current through it, electrons are emitted. A very high voltage is applied between
the cathode and the anode target as shown in Fig. 12.10. Depending upon the application, an X-ray
tube is designed with voltage in a high range from 20 kV to 20 MV. The electrons, accelerated by
the high voltage, impact the target producing X-rays. The target is known as focal point and is
usually made of a hard material like tungsten. Most of the kinetic energy of the impacting electrons
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Elements of Fracture Mechanics
is converted into heat. Thus, a proper cooling of the target is required. Also, the target cannot be
very small to avoid its overheating. The finite size of target plays an important role in affecting the
sharpness of imaging. This aspect will be discussed subsequently. A window of low absorbing
material (beryllium, aluminum) is provided for the exit of the X-rays from the tube. The X-rays
radiation spreads as shown in Fig. 12.11. X-rays pass through a work-piece to a recording film.
X-rays tube
Anode
Cathode
Focal point
High
voltage
Window
X-rays
Fig. 12.10 X-ray setup
Penetration Ability and Exposure
Absorption of X-rays is high in thick or dense materials. To have an appropriate intensity of X-rays
for a given material and a given thickness of a work-component, suitable anode-voltage is selected
for the X-rays tube. For example, X-rays generated by 150 kV can penetrate 30 mm thick steel plate
and 80 mm thick aluminum plate. Similar to a photographic camera, right exposure is given for
imaging defects. Too much or too little exposure will not give a proper contrast on the recording
film. Usually, an exposure is measured as E = At, where A is the filament current of cathode and t
the exposure time.
Loss of Sharpness of an Image
The image of a defect may not be sharp due to two considerations:
(i) Geometric unsharpness: The focal point works more like a point source from where X-rays
start spreading as shown in Fig. 12.11. However, as discussed earlier, it is difficult to have
the focal point very small; usually its size is between 2 mm ¥ 2 mm and 5 mm ¥ 5 mm. Due
to the finite size of the focal point, X-rays from the edge of a defect fall on a finite width of
the recording film, giving rise to a blurring of the edge (Fig. 12.12). The blurring can be
minimized by placing the focal point far away from the work-piece and the recording film
very close to the rear surface of the work-piece.
(ii) Effect on radiation on recording film: High density radiation interacts with the recording film
which has a thin layer of a silver halide as a main constituent to record the amount of
exposure. The radiographic exposure dislodges electrons from silver halide emulsion. These
electrons fly to cause ionization of adjacent silver halides causing blurring of the edges of a
defect. One can use low sensitivity film with a thinner spread of silver halide to suppress
Crack Detection through Non-Destructive Testing
261
Fig. 12.11 X-rays falling on a work-piece with a film placed
close to its rear surface
the blurring. But then, the exposure time on the film would have to be higher, thus creating
a condition of give and take.
12.5.3 Imaging through Gamma Rays
A gamma rays source is an unstable nucleus that continuously emits radiation similar to the
radiation of X-rays. Several kinds of gamma sources are used but the most common are Cobalt 60
and Iridium 192. The energy level of radiation is quite high, in the wide range of 100 keV to 1 MeV.
Because of their smaller wavelength, they are capable of penetrating dense and thick materials.
Sources of g -rays are supplied by a national atomic commission (i.e., Bhabha Atomic Reactor
Center, Mumbai in India). They are supplied in sealed capsules of dimensions ranging between
0.3 mm ¥ 0.3 mm and 6 mm ¥ 6 mm.
The gamma imaging of defects in a material is similar to X-ray imaging with associated
problems of contrast, blurring and optimum exposure. However, a gamma source is small and can
be inserted into a small cavity. For example, a gamma source can be placed within a pipe of small
internal diameter; the radiation sensitive film is then placed at the outside surface of the pipe.
A g -source continuously emits radiation and, thus, has a limited life, a few months to a few
years depending upon the type of g -rays source. It is worth noting that X-rays emission is
broadband, that is, radiation has a wide range of wavelength. In contrast, g -rays emission is
monochromatic. Also, in comparison to X-rays, the number of disintegration per unit time is much
less in g -rays and, therefore, substantially longer exposure time is required.
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Focal point
Defect
Geometric
unsharpness
Film
Fig. 12.12 The finite size of the focal point develops geometric
unsharpness on the film
12.5.4 Strong Points of Radiographic Imaging
∑
∑
∑
∑
Can detect defects in almost all materials including dense steel.
X-rays and g -rays can penetrate thick components and detect cracks.
No surface preparation of a work-piece required.
Sources of g -rays are very compact to use.
12.5.5 Limitations of Radiographic Imaging
∑ The technique is relatively expensive. Sophisticated equipment is required for X-rays and a
source of g - rays has a limited life.
∑ Access to both front and rear surfaces is needed.
∑ Very tight cracks (very small space between the crack faces) are difficult to be detected.
Also, minute discontinuities are not detected.
∑ Radiographic imaging gives no idea of how deep the defect is.
∑ Sharpness may be a concern in interpreting the images.
∑ A thin defect (e.g. a thin disk) normal to the propagation direction of X-rays or g -rays is not
recorded with good contrast as the progressive absorption through the defect is small
creating only small contrast.
∑ X-rays or g -rays are high energy electromagnetic waves and may cause injuries to workers.
Each worker in a radiographic unit has to carry a badge which progressively monitors the
radiation exposure to his body.
In short, radiographic imaging deals with powerful high frequency and very low wavelength
electromagnetic waves which pass through all kinds of materials. A defect absorbs the radiation
differentially and is recorded on a sensitive recording film with a different contrast.
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263
12.6 MAGNETIC PARTICLE INSPECTION
Magnetic particle inspection is capable of detecting surface and subsurface cracks in a work-piece
made of ferromagnetic materials such as most steels, nickel and cobalt. The technique is simple to
use, inexpensive, quick, sensitive, and reliable. However, inspection is limited to ferromagnetic
materials only.
12.6.1 Principle
A strong magnetic flux is generated within a work-piece. A defect disturbs the lines of force and
creates leakage of magnetic flux out of the surface of the work-piece. When minute magnetic
particles are sprayed, they get deposited on the surface at points which are close to the crack. This
happens because the disturbed magnetic flux passes through the particles. Figure 12.13 (a) shows
a crack open to the surface. Some magnetic particles may enter inside the cavity of the crack and
allow some magnetic flux pass through them. But most of the disturbed flux passes through a ridge
of particles which builds up at the mouth of the crack. It is worth noting that the width of the
particle ridge is much wider than the width of the crack mouth. Thus, the cracks can be observed
easily. This concept is similar to one used in liquid penetration test where the coloured penetrant
comes out of the crack cavity and spreads on the sides to make the crack visible.
A subsurface crack which is not too deep inside the surface (usually within 6 mm) is also
detectable. In Fig.13 (b), a fully embedded crack disturbs the magnetic flux and part of it tends to
leak out. When magnetic particles are sprayed, they get deposited on the surface to facilitate
magnetic lines of force pass through them. However, the deposition of magnetic particles due to a
subsurface crack is diffused whereas a surface crack shows clear and sharp deposition of particles.
(a)
(b)
Fig. 12.13 Magnetic flux through work-pieces, (a) a particle ridge is built up over the mouth
of a surface crack, and (b) magnetic particles are deposited due to a subsurface crack
12.6.2 Sensitivity
Magnetic particle inspection is quite sensitive and can detect cracks as small as 0.02 mm deep.
Also, it is capable of detecting surface cracks with opening as small as 0.002 mm (2 mm). However,
the ratio of the depth to width of the crack should be large (>5). A shallow crack with a wide
opening having depth/width ratio of the order of one does not really disturb the magnetic flux and
therefore magnetic particles do not get deposited.
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12.6.3 Hardware
The technique is versatile and many kinds of hardware models have been developed for different
geometry of work-pieces. A schematic diagram of a general purpose test setup is shown in
Fig. 12.14. Its flexible yokes can clamp a large variety of work-pieces. The coil develops a
longitudinal magnetic flux in the work-piece. The work-piece in the figure shows a corner crack
identified by magnetic particle inspection.
Fig. 12.14 A schematic diagram of a hardware setup with yokes
12.6.4 Flaw Orientation
Flaws normal to the magnetic flux direction disturb the magnetic lines of the force the most. On the
other hand, if a thin long crack is aligned to the magnetic flux direction, it would not be detected.
Usually, flaw orientations between 45° and 90° are detected. Thus, to detect all the cracks, the
work-piece should be tested in several orientations. For example, if defects are to be detected in a
tube, two kinds of magnetic flux should be generated. In Fig.12.15 (a), a coil is wound around the
tube generating longitudinal magnetic flux. The configuration detects cracks normal to the axes of
the tube. In Fig. 12.15 (b), a conductor rod is passed through the interior of the tube generating
circular flux. This kind of flux would detect cracks parallel to the axis of the tube.
Crack Detection through Non-Destructive Testing
(a)
265
(b)
Fig. 12.15 (a) Wound coils around a tube generating axial lines lines of force and
(b) a conducting rod inserted inside the tube generating circular flux
12.6.5 Magnetic Ink Powder
Particles used are minute in size (about 6 mm) and are made of a suitable ferromagnetic material
with good permeability like pulverized iron oxide (Fe3O4), carbonyl iron powder (pure iron), etc.
There are two ways of applying the magnetic particles, (i) wet and (ii) dry. In the wet route,
particles are suspended in water or oil and are applied on the component. In dry particle route, air
is the carrier which sprays particles on the surface.
Magnetic particles inspection can be performed even if there is a thin paint on the surface of the
work-piece and particles. Thus, particles are often coloured with black or red thin paint. Also, the
work-piece may be painted with a thin coat of white colour so that the build up of black or red
coloured particles on a defect is easily observed.
Fluorescent particles are also used, especially to detect hair line cracks. Fluorescent particles are
prepared by applying a thin coat of a fluorescent material on the surface of the ferromagnetic
particles. Then, the observation is made under dark conditions using UV light. In comparison to
crack detection through ordinary magnetic particles, the illumination of a crack through
fluorescent particles is vastly superior.
12.6.6 Voltage Source
Three kinds of voltage sources are used: (i) DC, (ii) AC, and (iii) Half Wave Rectification (HWAC).
A DC source gives full penetration and is suitable to detect both surface and sub-surface cracks. An
AC source works well with surface cracks. A HWAC source works well with dry particles as ripples
in magnetic flux keep on vibrating the particles making them more visible.
12.6.7 Demagnetization
Often, a work-piece needs to be demagnetized to avoid confusion in detecting defects. There are
several ways but one good method is to place the work-piece in reverse magnetic field and reduce
the magnitude gradually to zero.
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12.6.8 Strength and Limitations
The strength and limitations of the magnetic particle inspection are:
Strength
Simple to use.
Inexpensive.
Rugged equipments.
Sensitive enough to detect even hairline cracks.
No elaborate surface preparation of the work-piece required; only degreasing is adequate,
especially on machined and electroplated surfaces. However, rust, scale, etc., should be
removed.
∑ No elaborate safety required.
∑ Quite dependable if a component can be checked in several directions.
∑
∑
∑
∑
∑
Limitations
∑
∑
∑
∑
Only work-pieces made of ferromagnetic materials can be inspected.
Only surface or subsurface defects are detectable
Orientation of a crack with the direction of magnetic flux is important.
Often, demagnetization of a component is required.
In short, the magnetic particle inspection is a powerful technique to detect surface or subsurface
defects in ferromagnetic materials. The technique is inexpensive and simple to use.
12.7 CONCLUDING REMARKS
Fracture mechanics is applicable only when a crack exists in a work-piece. Real materials have
many defects like voids, inclusions, cracks, etc. But most of them are subcritical and safe. Reliable
non-destructive test methods are required to assure safety of components. These days, many nondestructive test techniques are available. However, only some of the widely used methods are
discussed in this chapter. Also, only the basics of various NDTs are covered in this chapter; for indepth details and various sophistications, readers are recommended to refer to advanced material.
Some time back, I met a German expert whose team developed an experimental test setup to
check a locomotive wheel for defects with ultrasonic testing at 21 locations. The testing of the
wheels is done at a fast pace so that all the wheels of a train rack can be tested within an hour. Such
preventive testing can save many human lives. Thus, the modern philosophy is to avoid, to the
maximum possible extent, a catastrophic failure by regularly checking the critical components
through non-destructive testing.
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REFERENCES
12.1 Mix P. E. (2005). Introduction to Nondestructive Testing, John Wiley & Sons, Inc., New Jersey.
12.2 Raj, Baldev, Jayakumar, T. and Thavasimuthu, M. (1997). Practical Nondestructive Testing,
Narosa Publishing House, New Delhi.
12.3 Boving, K. G. (1989). NDE Handbook, Butterworths, London.
12.4 Hemelrijck, D.V. and Anastassopoulos, A. (1996). Non Destructive Testing, A.A. Belkema,
Rotterdam.
12.5 Halmshaw, R. (1996). Introduction to the Non-Destructive Testing of Welded Joints, Woodhead
Publishing Ltd. Cambridge.
12.6 Cartz, L. (1995). Nondestructive Testing, ASM International, Material Park.
12.7 Bray, D.E. and Stanley R.K. (1989). Nondestructive Evaluation, McGraw-Hill Book Company,
New York.
12.8 http://www.ndt-ed.org/index_flash.htm
Index
12-node isoparametric 221
3-node triangle 221
3-point bend specimen 127
4-node quadrilateral 221
5% secant line 161
8-node isoparametric 221
A
Accelerated tests 210, 211
Acoustic emission 2
Airy Stress Function 56
Airy's Stress Function 45
Allowable stress 191
Aluminum alloy 196
Anastassopoulos, A. 267
Anderson, T. L. 34
Anderson, T.L. 8
Anelastic deformation 99
Anelastic deformation 14, 24, 25
Angular probe 259
Anodic dissolution 202
Arc-Shaped Tension (AT) Specimen 94
Ashby, M.F. 216
ASTM Code Designation E 399-83 156
AT specimen 156
Austenitic stainless steel 196
Axially Cracked Pressurized Cylinder 143
B
Barsom, J.M. 8, 34, 216
Barsoum Element 230
Barsoum, R.S. 232
Bauschinger effect 36
Bauschinger’s effect 118
Begley 129
Bending Moment on a Centre-Cracked Plate
Biharmonic Equation 45, 47
Blunting line 169
Boving, K. G. 267
Bowie, O.L. 232
Bowles 153
Bray, D.E. 267
Brittle failure 4
Brittle material 35, 161
Broek, D. 7, 34, 216, 247
Broek, David 217
Brown Jr. W.F. 232
Buckling 2
Burrs 115
C
C-Scan 258
Cartz, L. 267
Cast Iron 196
Catastrophic failure 27
Cathode 204
Cauchy–Riemann relations 46
Centre-Cracked Plate 90, 139
Centre-cracked Plate 85
Chan, S. K. 232
Chevron notch 159
Circumferentially Cracked Cylinder 144
Circumferentially Cracked Round Bar 91
Clapeyron’s Theorem 20
Cleavage failure 4
Clip Gauge 160
Cobalt 60 261
Coin test 250
Collapse-load 132
82
270
Index
Collinear Cracks 64
Comet Jet airplanes 189
Comet Jetliner 4
Compact Tension Specimen 140
Compatibility 42
Compatibility Relations 42
Complex variables 45
Compliance 16
Constant amplitude cyclic load 196
Constant amplitude fatigue load 198
Constant amplitude load 189
Constant amplitude loading 191
Constant displacement method 205, 208
Constant Load 17
Constant Load Method 205
Constant load test 206
Constraints on the specimen size 169
Cook 232
Corner Cracks 75
Corrosion 203
Corrosion chamber 205
Corrosion mechanisms 203
Corrosive environment 188, 204, 205
Corrosive environments 203
Cottrell 149, 153
Crack Closure 199
Crack closure 213
Crack growth per unit cycle 192
Crack growth rate 207
Crack Initiation 192
Crack Mouth Opening Displacement
(CMOD) 160, 161, 179
Crack nucleation 194
Crack opening displacement (COD) 53
Crack propagation rate 198
Crack resistance 14
Crack tip opening displacement (CTOD)
4, 53, 149
Critical crack length 13
Critical CTOD 179
Critical energy release rate 30
Critical J-Integral 128
Critical stress intensity factor 156, 197
CT specimen 157
CTODc 150, 179
D
Damage tolerance design
DCB specimen 21
DCT specimen 156
196
Demagnetization 265
Developer 251
Diamond 100
Diffusion 204
Disc-shaped Compact Tension (DCT)
Specimen 95
Dislocations 202, 203
Dissolution rate 204
Divergence Theorem 123
Double cantilever beam (DCB) 16, 121, 172
Double-Edge Notched Plate 142
Ductile fracture 4
Dugdale 108
Dugdale approach 108, 151
Dwell time 252
Dye penetrants 169
Dye-penetration 2
E
Echo 254
Edge crack 196
Effective crack 150
Effective Crack Length 104
Elastic-plastic behavior 118
Elastic-plastic fracture 198
Elastic-plastic fracture mechanics (EPFM) 36, 119
Elber 201, 216
Electrolytic cell 204
Element stiffness matrix 221
Elliptical Crack 73
Elliptical integral of a second kind 74
Ellyin, Fernard 216
Endoscope 249
Endurance limit 3, 191
Energy release rate 4, 14, 15, 28, 36, 227
Engle, R.M. 216
Environment-Assisted Fatigue Failure 211
Environment-Assisted Fracture 202, 203, 209
Environmental degradation 1, 27
Equilibrium equations 42
Erdogan 197, 216, 247
Extrusions 194
F
Factor of safety 2
Far field stress 36, 37, 38
Fatigue 1, 2, 27
Fatigue Crack Growth 159, 213
Fatigue failure 188, 211, 213
Fatigue life 191
Index
Fatigue load 194
FEM 218
Ferrite-Pearlite steel 195
Field equations 42
Filament current 260
Finite element analysis 2
Finite element method 218
Fixed Grip 18, 20
Flow stress 168
Fluctuating loads 2, 189
Fluorescent particles 265
Forman 197, 216
Fractured surface 114, 233, 234
Frequency 212
G
g -rays 259
Gamma Rays 261
Gauss points 226
Gdoutos, E.E. 34, 216
Geometric unsharpness 260
George Irwin 4
Gifford, Jr. L. N. 232
Glide bands 194
Global stiffness matrix 221
Goodman, J. 8
Grain boundaries 203, 204
Green's functions 63
Griffith, A.A. 4, 8, 9, 10, 14, 34
H
Half Wave Rectification 265
Halmshaw, R. 267
Hearing 250
Heat treatment 209
Hemelrijck, D.V. 267
Hertzberg, R. W. 34, 217
Hilton, P. D. 232
Humidity 203, 209
Hydrogen atoms 204
Hydrogen embrittlement 204, 213
I
Immersion inspection 258
Incubation time 203, 206
Inglis, C. E. 24, 34
Initial stress intensity factor 206
Initiation life 190, 193, 194
Intense plastic deformation 169
Intergranular 4
Intergranular crack growth 204
Interlaminar cracks 250
Interlaminar GIc 172
Interlaminar GIIc 177
Internal Pressure 67
Interpolation functions 219
Intrusions 194
Iridium 192 261
Irwin, G.R. 8, 14, 34, 39, 107
Irwin Plastic Zone Correction 105
Irwin's correction 150
Isida, M. 232
Izod notch sensitivity test 154
J
J-Integral 4, 119, 151, 226
Janssen, M. 34, 217
Jayakumar, T. 267
JIc-test 164
Jones D.R.H 216
K
Kaesche, Helmut 217
Kanninen, M.F. 34
Kearney, V.E. 216
KIc-Test Technique 156
Kies 39
Knott, J.F. 34, 217
L
Lapped Joint 93
leak before break 250
Leonardo da Vinci 3
Liberty ships 3, 5
Linear Elastic Fracture Mechanics
(LEFM) 35, 99, 156
Liquid Metal Embrittlement 210
Liquid Penetration Inspection 251
Liquid penetration method 249
Liu, A. 8
Longitudinal waves 254
Looked 129
M
Maganetic Particle Inspection 263
Magnetic flux 263
Magnetic flux method 2
Magnetic Ink Powder 265
Magnetic particle inspection 249
Malkus 232
271
272
Index
Martensitic steel 195
Maximum Tangential Stress (MTS)
Criterion 235, 236
McNicol, R.C. 216
Mean stress 189, 212
Meguid, S.A. 34, 216
Micro-cracks 203
Micromechanisms 203
Microvoids 5
Mises 1, 3
Mises criterion 101
Mix, P. E. 267
Mixed mode loading 233
Mode I 5, 6
Mode II 5
Mode III 5
Modified Griffith Criterion 235
MTS criterion 237
Muskhelishvili 70
N
NDT 248
Non-destructive techniques 213
Non-destructive tests 2
Non-linear elastic 118
Nuclear Reactor Steel 108
O
Oblique Edge Crack 93
Onset of the crack growth 127, 167
Orowan 14, 34
Over-aging 209
Overload 198
Overload pulse 213
P
Paris law 195, 197, 201, 213
Passive layer 203, 204, 211
Path Independence 122
Penetrant liquid 251
Perspex (Plexiglass) 108
Piezoelectric material 256
Plane strain 28, 105, 112, 205
Plane stress 28, 105, 112, 206
Plastic deformation 5, 15
Plastic zone 24, 36, 100
Plastic zone shape 101
Plastic zone size 25, 28, 102, 108, 198
Plesha 232
Pook, L.P. 216
Pop-in 161
Popelar, C. H. 34
Potential energy 15, 218, 221
Primary waves 254
Principle of Superposition 66
Probe 255
Propagation life 190
Propagation time 213
Pulse-echo instrument 255
Q
Quarter-Point element
Quartz 256
230
R
R-curve 25, 27, 29
Radiation exposure 262
Radiation filming 2
Radiographic imaging 249, 259
Radius of curvature 193
Raghavan, V. 217
Raj, Baldev 267
Ramberg–Osgood equation 125
Ramesh, K. 8, 217
Ratwani 197, 216
Reddy 232
Residual stresses 198
Rice 4, 8, 119
Rolfe, S.T. 8, 34, 216
Rooke 153
Root mean square value 201
S
S-N curve 191, 197
S-N diagram 3
Salt concentration 209
Sanford, R. J. 34
Schijve 201, 216
Second phase particles 5
Secondary waves 254
SED criterion 241
Self-similar plane 159
Semi-elliptical Cracks 74
SENB specimen 156, 166, 179
Shape functions 218
Shear waves 254
Sigmoidal Curve 192
Sih, G. C. 247
Single-Edge-Cracked Plate 90
Index
Single-Edge-Notch-Bend (SENB) Specimen 89
Singular Element Method 228
Singular finite elements 228
Singularity 41
Slip bands 203
Smell 250
Smith, R.A. 216
Solution 86
Specimen-Dimensions 158
Square root singularity 38
Srawley, J.E. 232
Stable crack growth 26
Stanley R.K. 267
Stiffness Derivative Method 228
Strain energy 15
Strain energy density 226
Strain Energy Density (SED) Criterion 235, 240
Strain rate 5
Strain-displacement relations 42
Stress intensity factor 4, 14, 35, 39, 203
Stress ratio 190, 193, 195
Stress waves 253
Stress-Strain Relations 43
Striation 200
Suresh, S. 216
Surface crack 74
Surface energy 24
T
Taylor, R. L. 232
Tension-compression fatigue 190
Tension-tension fatigue 190
Thavasimuthu, M. 267
Thick plate 5, 28, 38
Thin plates 5, 28, 38
Three-Point Bend Specimen 138
Threshold SIF 205
Time-to-failure 205
Titanium alloy 108
Transgranular 4
Transitional behaviour 112
Transmission losses 257
Transmitter-receiver transducers 258
Tresca 1, 3
Tresca yield criterion 102
Trial functions 219
Triangular element 219
Tuba, I.S. 232
U
Ultimate tensile strength 168
Ultrasonic crack detection 2
Ultrasonic testing 249, 253
Uncracked ligament 128
UV light 265
V
V-notch 159, 179
Variable amplitude fatigue load 201
Variable amplitude load 189
Variational method 218
Various Test Specimens 156
Virtual work principle 221
W
Wanhill, R.J.H. 34, 217
Watwood Jr., V. B. 232
Wedge Loads 62
Wells 4, 8, 149, 153
Westergaard 45
Westergaard approach 58
Westergaard function 50
Wilson, W.K. 232
Wöhler, A. 8
Work-hardening 203, 210
Wu, E.M. 247
X
X-ray
2, 259
Y
Yielding 1
Z
Zienkiewicz 232
Zuidema, J. 34, 217
273
Author’s Profile
Prof. Prashant Kumar was till recently with the Department of Mechanical Engineering, Indian
Institute of Technology, Kanpur. He has over three decades of teaching and research experience. A
B. Tech. in Mechanical Engineering from IIT Kanpur, he has M.S. in Design from University of
California, Berkeley, USA. He worked as a Design Engineer in Los Angeles for over two years. In
1976, he obtained his Ph. D in Mechanics of Solids from Brown University, USA.
Prof. Prashant Kumar has published a number of papers in international and national journals
and magazines. He has worked on sponsored projects on fracture aspects of polymer composites.
He has been a consultant to many industries around Kanpur and national organizations like HAL
Korwa, ADE Bangalore and ADA Bangalore.
Prof. Prashant Kumar was the founder Head of Department of newly created Design Programme
at IIT Kanpur from year 2002 to 2005. Earlier, he was Head of Department of Mechanical Engineering
from year 1998 to 2001.